Angle bisector theorem examples

WEBVTT
Kind: captions
Language: en

00:00:00.499 --> 00:00:02.810
I thought I would do a few
examples using the angle

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bisector theorem.

00:00:03.950 --> 00:00:05.700
So in this first
triangle right over here,

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we're given that this
side has length 3,

00:00:08.440 --> 00:00:10.330
this side has length 6.

00:00:10.330 --> 00:00:12.030
And this little
dotted line here,

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this is clearly
the angle bisector,

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because they're telling
us that this angle is

00:00:15.444 --> 00:00:17.153
congruent to that
angle right over there.

00:00:17.153 --> 00:00:18.693
And then they tell
us that the length

00:00:18.693 --> 00:00:21.220
of just this part of this
side right over here is 2.

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So from here to here is 2.

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And that this length is x.

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So let's figure out what x is.

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So the angle bisector
theorem tells us

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that the ratio of 3 to 2 is
going to be equal to 6 to x.

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And then we can
just solve for x.

00:00:35.240 --> 00:00:40.290
So 3 to 2 is going to
be equal to 6 to x.

00:00:40.290 --> 00:00:42.770
And then once again, you
could just cross multiply,

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or you could multiply
both sides by 2 and x.

00:00:45.290 --> 00:00:47.000
That kind of gives
you the same result.

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If you cross multiply, you get
3x is equal to 2 times 6 is 12.

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x is equal to, divide both
sides by 3, x is equal to 4.

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So in this case,
x is equal to 4.

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And this is kind of interesting,
because we just realized now

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that this side, this entire
side right over here,

00:01:05.069 --> 00:01:06.615
is going to be equal to 6.

00:01:06.615 --> 00:01:08.240
So even though it
doesn't look that way

00:01:08.240 --> 00:01:11.885
based on how it's drawn, this is
actually an isosceles triangle

00:01:11.885 --> 00:01:16.000
that has a 6 and a 6, and then
the base right over here is 3.

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It's kind of interesting.

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Over here we're given that this
length is 5, this length is 7,

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this entire side is 10.

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And then we have this angle
bisector right over there.

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And we need to figure out just
this part of the triangle,

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between this point, if
we call this point A,

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and this point right over here.

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We need to find the length
of AB right over here.

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So once again, angle bisector
theorem, the ratio of 5

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to this, let me do this in a
new color, the ratio of 5 to x

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is going to be equal
to the ratio of 7

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to this distance
right over here.

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And what is that distance?

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Well, if the whole thing
is 10, and this is x,

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then this distance right over
here is going to be 10 minus x.

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So the ratio of 5 to x is
equal to 7 over 10 minus x.

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And we can cross
multiply 5 times 10

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minus x is 50 minus 5x.

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And then x times
7 is equal to 7x.

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Add 5x to both sides
of this equation,

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you get 50 is equal to 12x.

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We can divide both sides by
12, and we get 50 over 12

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is equal to x.

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And we can reduce this.

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Let's see if you divide the
numerator and denominator by 2,

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you get this is the
same thing as 25 over 6,

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which is the same thing, if
we want to write it as a mixed

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number, as 4, 24
over 6 is 4, and then

00:02:45.620 --> 00:02:47.300
you have 1/6 left over.

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4 and 1/6.

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So this length right
over here is going,

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oh sorry, this length right
over here, x is 4 and 1/6.

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And then this
length over here is

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going to be 10 minus 4 and 1/6.

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What is that?

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5 and 5/6.


Inradius, perimeter, and area

WEBVTT
Kind: captions
Language: en

00:00:00.570 --> 00:00:06.930
We're told the triangle ABC
has perimeter P and inradius r

00:00:06.930 --> 00:00:09.590
and then they want us
to find the area of ABC

00:00:09.590 --> 00:00:11.129
in terms of P and r.

00:00:11.129 --> 00:00:12.670
So we know that the
perimeter is just

00:00:12.670 --> 00:00:14.900
the sum of the sides
of the triangle,

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or how long a fence
would have to be

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if you wanted to go
around the triangle.

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And let's just remind
ourselves what the inradius is.

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If we take the angle
bisectors of each of these

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vertices-- each of these
angles right over here.

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So bisect that right over
there and then bisect

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that right over there.

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This angle is going to
be equal to that angle.

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This angle is going to
be equal to that angle

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and then this angle is going to
be equal to that angle there.

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And the point where those angle
bisectors intersect, that right

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over there, is our
incenter and it

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is equidistant from
all of the three sides.

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And the distance from those
sides, that's the inradius.

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So let me draw the inradius.

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So when you find the distance
between a point and a line,

00:01:01.210 --> 00:01:02.584
you want to drop
a perpendicular.

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So this length right over
here is the inradius.

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This length right over
here is the inradius

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and this length right
over here is the inradius.

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And if you want, you could
draw an incircle here

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with the center at the
incenter and with the radius r

00:01:18.330 --> 00:01:20.450
and that circle would
look something like this.

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We don't have to necessarily
draw it for this problem.

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So you could draw a circle
that looks something like that.

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And then we'd call
that the incircle.

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So let's think about how we can
find the area here, especially

00:01:30.850 --> 00:01:32.886
in terms of this inradius.

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Well, the cool thing
about the inradius

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is it looks like the
altitude-- or this looks

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like the altitude for this
triangle right over here,

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triangle A. Let's
label the center.

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Let's call it I for incenter.

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This r right over here is
the altitude of triangle AIC.

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This r is the altitude
of triangle BIC.

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And this r, which we didn't
label, that r right over there

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is the altitude of triangle AIB.

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And we know-- and
so we could find

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the area of each
of those triangles

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in terms of both
r and their bases

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and maybe if we sum up the
area of all the triangles,

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we can get something in terms of
our perimeter and our inradius.

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So let's just try to do this.

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So the area of the entire
triangle, the area of ABC,

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is going to be
equal to-- and I'll

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color code this-- is going to
be equal to the area of AIC.

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So that's what I'm
shading here in magenta.

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It's going to be equal to
the area of AIC plus the area

00:02:34.750 --> 00:02:37.551
of BIC, which is this
triangle right over here.

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Actually let me do that
in a different color.

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I've already used the blue.

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So let me do that in orange.

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Plus the area of BIC.

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So that's this area
right over here.

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And then finally plus the area--
I'll do this in a, let's see,

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I'll use this pink color--
plus the area of AIB.

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That is the area AIB.

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Take the sum of the areas
of these two triangles,

00:03:13.280 --> 00:03:15.590
you got the area of
the larger triangle.

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Now AIC, the area
of AIC, is going

00:03:19.030 --> 00:03:21.700
to be equal to 1/2
base times height.

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So this is going to be 1/2.

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The base is the length
of AC, 1/2 AC times

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the height-- times this
altitude right over here,

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which is just going
to be r-- times r.

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That's the area of AIC.

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And then the area of BIC is
going to be 1/2 the base,

00:03:41.280 --> 00:03:45.540
which is BC, times the
height, which is r.

00:03:45.540 --> 00:03:49.260
And then plus the area of
AIB, this one over here,

00:03:49.260 --> 00:03:51.620
is going to be 1/2
the base, which

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is the length of this side
AB, times the height, which

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is once again r.

00:04:00.090 --> 00:04:03.900
And over here, we can factor out
a 1/2 r from all of these terms

00:04:03.900 --> 00:04:16.233
and you get 1/2 r times
AC plus BC plus AB.

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And I think you see
where this is going.

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Plus-- now that's a different
shade of pink-- plus AB.

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Now what is AC plus BC plus AB?

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Well that's going to
be the perimeter, P,

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if you just take the
sum of the sides.

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So that is the perimeter of P
and it looks like we're done.

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The area of our triangle ABC
is equal to 1/2 times r times

00:04:51.680 --> 00:04:54.750
the perimeter, which is
kind of a neat result.

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1/2 times the inradius times
the perimeter of the triangle.

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Or sometimes you'll see
it written like this.

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It's equal to r times P
over s-- sorry, P over 2.

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And this term right over here,
the perimeter divided by 2,

00:05:10.290 --> 00:05:11.873
is sometimes called
the semiperimeter.

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And sometimes it's denoted
by s so sometimes you'll

00:05:19.940 --> 00:05:22.760
see the area is
equal to r times s,

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where s is the semiperimeter.

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It's the perimeter divided by 2.

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I personally like it
this way a little bit

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more because I remember
that P is perimeter.

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This is useful because obviously
now if someone gives you

00:05:33.430 --> 00:05:35.400
an inradius and a perimeter,
you can figure out

00:05:35.400 --> 00:05:36.640
the area of a triangle.

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Or if someone gives you
the area of the triangle

00:05:38.680 --> 00:05:40.513
and the perimeter, you
can get the inradius.

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So if they give you
two of these variables,

00:05:42.630 --> 00:05:44.140
you can always get the third.

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So for example, if this was
a triangle right over here,

00:05:47.940 --> 00:05:50.750
this is maybe the most famous
of the right triangles.

00:05:50.750 --> 00:05:55.235
If I have a triangle that
has lengths 3, 4, and 5,

00:05:55.235 --> 00:05:56.610
we know this is
a right triangle.

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You can verify this from
the Pythagorean theorem.

00:05:58.674 --> 00:06:00.090
And if someone
were to say what is

00:06:00.090 --> 00:06:03.420
the inradius of this
triangle right over here?

00:06:03.420 --> 00:06:05.626
Well we can figure out
the area pretty easily.

00:06:05.626 --> 00:06:07.000
We know this is
a right triangle.

00:06:07.000 --> 00:06:09.800
3 squared plus 4 squared
is equal to 5 squared.

00:06:09.800 --> 00:06:16.370
So the area is going to be
equal to 3 times 4 times 1/2.

00:06:16.370 --> 00:06:19.040
So 3 times 4 times
1/2 is 6 and then

00:06:19.040 --> 00:06:21.170
the perimeter here
is going to be

00:06:21.170 --> 00:06:26.690
equal to 3 plus 4, which
is 7, plus 5 is 12.

00:06:26.690 --> 00:06:29.670
And so we have the area.

00:06:29.670 --> 00:06:35.720
So let's write this area is
equal to 1/2 times the inradius

00:06:35.720 --> 00:06:37.510
times the perimeter.

00:06:37.510 --> 00:06:42.770
So here we have 12 is equal
to 1/2 times the inradius

00:06:42.770 --> 00:06:44.750
times the perimeter.

00:06:44.750 --> 00:06:46.750
So we have-- oh
sorry, we have 6.

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Let me write this in.

00:06:47.720 --> 00:06:49.620
The area is 6.

00:06:49.620 --> 00:06:55.200
We have 6 is equal to 1/2
times the inradius times 12.

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And so in this situation,
1/2 times 12 is just 6.

00:06:58.170 --> 00:07:00.310
We have 6 is equal to 6r.

00:07:00.310 --> 00:07:03.560
Divide both sides by 6,
you get r is equal to 1.

00:07:03.560 --> 00:07:06.440
So if you were to draw
the inradius for this one,

00:07:06.440 --> 00:07:08.120
which is kind of a neat result.

00:07:08.120 --> 00:07:11.245
So let me draw some
angle bisectors here.

00:07:13.830 --> 00:07:17.570
This 3, 4, 5 right triangle
has an inradius of 1.

00:07:17.570 --> 00:07:19.980
So this distance
equals this distance,

00:07:19.980 --> 00:07:22.530
which is equal to
this distance, which

00:07:22.530 --> 00:07:28.290
is equal to 1, which is
kind of a neat result.


Angle bisector theorem proof

WEBVTT
Kind: captions
Language: en

00:00:00.490 --> 00:00:01.990
What I want to do
first is just show

00:00:01.990 --> 00:00:05.100
you what the angle
bisector theorem is

00:00:05.100 --> 00:00:07.520
and then we'll actually
prove it for ourselves.

00:00:07.520 --> 00:00:10.490
So I just have an
arbitrary triangle right

00:00:10.490 --> 00:00:12.180
over here, triangle ABC.

00:00:12.180 --> 00:00:13.750
And what I'm going
to do is I'm going

00:00:13.750 --> 00:00:17.249
to draw an angle bisector
for this angle up here.

00:00:17.249 --> 00:00:19.540
And we could have done it
with any of the three angles,

00:00:19.540 --> 00:00:20.270
but I'll just do this one.

00:00:20.270 --> 00:00:22.290
I'll make our proof
a little bit easier.

00:00:22.290 --> 00:00:25.670
So I'm just going to bisect
this angle, angle ABC.

00:00:25.670 --> 00:00:29.900
So let's just say that's the
angle bisector of angle ABC,

00:00:29.900 --> 00:00:31.950
and so this angle
right over here

00:00:31.950 --> 00:00:35.040
is equal to this
angle right over here.

00:00:35.040 --> 00:00:38.700
And let me call this point down
here-- let me call it point D.

00:00:38.700 --> 00:00:47.560
The angle bisector
theorem tells us

00:00:47.560 --> 00:00:50.290
that the ratio between
the sides that aren't this

00:00:50.290 --> 00:00:52.680
bisector-- so when I put
this angle bisector here,

00:00:52.680 --> 00:00:55.750
it created two smaller triangles
out of that larger one.

00:00:55.750 --> 00:00:57.410
The angle bisector
theorem tells us

00:00:57.410 --> 00:01:01.360
the ratios between the other
sides of these two triangles

00:01:01.360 --> 00:01:04.090
that we've now created
are going to be the same.

00:01:04.090 --> 00:01:12.620
So it tells us that
the ratio of AB to AD

00:01:12.620 --> 00:01:19.310
is going to be equal to the
ratio of BC to, you could say,

00:01:19.310 --> 00:01:21.820
CD.

00:01:21.820 --> 00:01:24.820
So the ratio of--
I'll color code it.

00:01:24.820 --> 00:01:32.310
The ratio of that,
which is this, to this

00:01:32.310 --> 00:01:36.590
is going to be equal to
the ratio of this, which

00:01:36.590 --> 00:01:40.970
is that, to this right
over here-- to CD,

00:01:40.970 --> 00:01:42.020
which is that over here.

00:01:42.020 --> 00:01:43.799
So once you see the
ratio of that to that,

00:01:43.799 --> 00:01:46.090
it's going to be the same as
the ratio of that to that.

00:01:46.090 --> 00:01:48.214
So that's kind of a cool
result, but you can't just

00:01:48.214 --> 00:01:50.680
accept it on faith because
it's a cool result.

00:01:50.680 --> 00:01:52.900
You want to prove
it to ourselves.

00:01:52.900 --> 00:01:54.620
And so you can imagine
right over here,

00:01:54.620 --> 00:01:56.140
we have some ratios set up.

00:01:56.140 --> 00:01:58.920
So we're going to prove it
using similar triangles.

00:01:58.920 --> 00:02:02.120
And unfortunate for us, these
two triangles right here

00:02:02.120 --> 00:02:04.599
aren't necessarily similar.

00:02:04.599 --> 00:02:07.015
We know that these two angles
are congruent to each other,

00:02:07.015 --> 00:02:08.490
but we don't know
whether this angle is

00:02:08.490 --> 00:02:09.735
equal to that angle
or that angle.

00:02:09.735 --> 00:02:10.318
We don't know.

00:02:10.318 --> 00:02:12.240
We can't make any
statements like that.

00:02:12.240 --> 00:02:14.690
So in order to actually set
up this type of a statement,

00:02:14.690 --> 00:02:17.310
we'll have to construct
maybe another triangle that

00:02:17.310 --> 00:02:20.440
will be similar to one
of these right over here.

00:02:20.440 --> 00:02:27.060
And one way to do it would
be to draw another line.

00:02:27.060 --> 00:02:29.545
And this proof
wasn't obvious to me

00:02:29.545 --> 00:02:31.170
the first time that
I thought about it,

00:02:31.170 --> 00:02:33.350
so don't worry if it's
not obvious to you.

00:02:33.350 --> 00:02:37.130
What happens is if we can
continue this bisector--

00:02:37.130 --> 00:02:38.830
this angle bisector
right over here,

00:02:38.830 --> 00:02:40.030
so let's just continue it.

00:02:40.030 --> 00:02:42.970
It just keeps going
on and on and on.

00:02:42.970 --> 00:02:45.860
And let's also-- maybe we can
construct a similar triangle

00:02:45.860 --> 00:02:47.910
to this triangle
over here if we draw

00:02:47.910 --> 00:02:51.660
a line that's parallel
to AB down here.

00:02:51.660 --> 00:02:53.070
So let's try to do that.

00:02:53.070 --> 00:02:57.070
So I'm just going to say,
well, if C is not on AB,

00:02:57.070 --> 00:03:00.080
you could always find
a point or a line that

00:03:00.080 --> 00:03:02.920
goes through C that
is parallel to AB.

00:03:02.920 --> 00:03:05.500
So by definition, let's
just create another line

00:03:05.500 --> 00:03:07.190
right over here.

00:03:07.190 --> 00:03:10.770
And let's call this
point right over here F

00:03:10.770 --> 00:03:13.990
and let's just pick
this line in such a way

00:03:13.990 --> 00:03:16.570
that FC is parallel to AB.

00:03:16.570 --> 00:03:21.120
So this is parallel to
that right over there.

00:03:24.520 --> 00:03:26.550
And we could just
construct it that way.

00:03:26.550 --> 00:03:28.240
And now we have some
interesting things.

00:03:28.240 --> 00:03:29.614
And we did it that
way so that we

00:03:29.614 --> 00:03:32.680
can make these two triangles
be similar to each other.

00:03:32.680 --> 00:03:33.622
So let's see that.

00:03:33.622 --> 00:03:34.580
Let's see what happens.

00:03:34.580 --> 00:03:36.288
So before we even
think about similarity,

00:03:36.288 --> 00:03:40.640
let's think about what we know
about some of the angles here.

00:03:40.640 --> 00:03:44.160
We know that we have
alternate interior angles-- so

00:03:44.160 --> 00:03:45.910
just think about these
two parallel lines.

00:03:45.910 --> 00:03:48.790
So I could imagine AB
keeps going like that.

00:03:48.790 --> 00:03:50.840
FC keeps going like that.

00:03:50.840 --> 00:03:54.000
And line BD right
here is a transversal.

00:03:54.000 --> 00:03:57.580
Then whatever this
angle is, this angle

00:03:57.580 --> 00:04:00.226
is going to be as well, from
alternate interior angles,

00:04:00.226 --> 00:04:02.350
which we've talked a lot
about when we first talked

00:04:02.350 --> 00:04:05.330
about angles with
transversals and all of that.

00:04:05.330 --> 00:04:07.540
So these two angles are
going to be the same.

00:04:07.540 --> 00:04:09.160
But this angle and
this angle are also

00:04:09.160 --> 00:04:11.250
going to be the same, because
this angle and that angle

00:04:11.250 --> 00:04:11.791
are the same.

00:04:11.791 --> 00:04:12.880
This is a bisector.

00:04:12.880 --> 00:04:14.790
Because this is a
bisector, we know

00:04:14.790 --> 00:04:18.790
that angle ABD is the
same as angle DBC.

00:04:18.790 --> 00:04:20.730
So whatever this angle
is, that angle is.

00:04:20.730 --> 00:04:22.600
And so is this angle.

00:04:22.600 --> 00:04:25.490
And that gives us kind
of an interesting result,

00:04:25.490 --> 00:04:29.180
because here we have
a situation where

00:04:29.180 --> 00:04:32.400
if you look at this
larger triangle BFC,

00:04:32.400 --> 00:04:34.990
we have two base angles
that are the same, which

00:04:34.990 --> 00:04:37.290
means this must be an
isosceles triangle.

00:04:37.290 --> 00:04:39.925
So BC must be the same as FC.

00:04:43.510 --> 00:04:44.670
So that was kind of cool.

00:04:44.670 --> 00:04:47.341
We just used the transversal and
the alternate interior angles

00:04:47.341 --> 00:04:48.715
to show that these
are isosceles,

00:04:48.715 --> 00:04:51.690
and that BC and FC
are the same thing.

00:04:51.690 --> 00:04:56.460
And that could be
useful, because we

00:04:56.460 --> 00:04:58.890
have a feeling that this
triangle and this triangle

00:04:58.890 --> 00:04:59.890
are going to be similar.

00:04:59.890 --> 00:05:01.050
We haven't proven it yet.

00:05:01.050 --> 00:05:04.520
But how will that help us get
something about BC up here?

00:05:04.520 --> 00:05:08.110
But we just showed that BC
and FC are the same thing.

00:05:08.110 --> 00:05:09.730
So this is going to
be the same thing.

00:05:09.730 --> 00:05:11.140
If we want to
prove it, if we can

00:05:11.140 --> 00:05:14.420
prove that the ratio of
AB to AD is the same thing

00:05:14.420 --> 00:05:17.240
as the ratio of FC
to CD, we're going

00:05:17.240 --> 00:05:22.240
to be there because BC, we
just showed, is equal to FC.

00:05:22.240 --> 00:05:24.030
But let's not start
with the theorem.

00:05:24.030 --> 00:05:25.446
Let's actually get
to the theorem.

00:05:25.446 --> 00:05:27.290
So FC is parallel
to AB, [? able ?]

00:05:27.290 --> 00:05:30.910
to set up this one
isosceles triangle,

00:05:30.910 --> 00:05:32.289
so these sides are congruent.

00:05:32.289 --> 00:05:34.080
Now, let's look at some
of the other angles

00:05:34.080 --> 00:05:37.180
here and make ourselves
feel good about it.

00:05:37.180 --> 00:05:38.320
Well, we have this.

00:05:38.320 --> 00:05:41.450
If we look at triangle ABD, so
this triangle right over here,

00:05:41.450 --> 00:05:44.160
and triangle FDC, we
already established

00:05:44.160 --> 00:05:46.930
that they have one set of
angles that are the same.

00:05:46.930 --> 00:05:49.310
And then, and then
they also both-- ABD

00:05:49.310 --> 00:05:51.460
has this angle right
over here, which

00:05:51.460 --> 00:05:53.630
is a vertical angle
with this one over here,

00:05:53.630 --> 00:05:54.880
so they're congruent.

00:05:54.880 --> 00:05:57.490
And we know if two triangles
have two angles that

00:05:57.490 --> 00:05:59.170
are the same, actually
the third one's

00:05:59.170 --> 00:06:00.380
going to be the same as well.

00:06:00.380 --> 00:06:02.560
Or you could say by the
angle-angle similarity

00:06:02.560 --> 00:06:05.917
postulate, these two
triangles are similar.

00:06:05.917 --> 00:06:07.000
So let me write that down.

00:06:07.000 --> 00:06:09.650
You want to make sure you get
the corresponding sides right.

00:06:09.650 --> 00:06:11.382
We now know by
angle-angle-- and I'm

00:06:11.382 --> 00:06:12.840
going to start at
the green angle--

00:06:12.840 --> 00:06:18.530
that triangle B-- and
then the blue angle-- BDA

00:06:18.530 --> 00:06:23.130
is similar to triangle--
so then once again,

00:06:23.130 --> 00:06:25.270
let's start with the
green angle, F. Then,

00:06:25.270 --> 00:06:26.652
you go to the blue angle, FDC.

00:06:31.030 --> 00:06:33.640
And here, we want to eventually
get to the angle bisector

00:06:33.640 --> 00:06:39.090
theorem, so we want to look at
the ratio between AB and AD.

00:06:39.090 --> 00:06:40.740
Similar triangles,
either you could

00:06:40.740 --> 00:06:42.820
find the ratio between
corresponding sides

00:06:42.820 --> 00:06:45.890
are going to be
similar triangles,

00:06:45.890 --> 00:06:48.040
or you could find
the ratio between two

00:06:48.040 --> 00:06:50.630
sides of a similar triangle
and compare them to the ratio

00:06:50.630 --> 00:06:53.820
the same two corresponding sides
on the other similar triangle,

00:06:53.820 --> 00:06:55.052
and they should be the same.

00:06:55.052 --> 00:06:57.010
So by similar triangles,
we know that the ratio

00:06:57.010 --> 00:07:03.130
of AB-- and this, by the way,
was by angle-angle similarity.

00:07:03.130 --> 00:07:04.980
Want to write that down.

00:07:04.980 --> 00:07:06.550
So now that we know
they're similar,

00:07:06.550 --> 00:07:12.717
we know the ratio of AB to
AD is going to be equal to--

00:07:12.717 --> 00:07:15.050
and we could even look here
for the corresponding sides.

00:07:15.050 --> 00:07:17.610
The ratio of AB, the
corresponding side

00:07:17.610 --> 00:07:23.553
is going to be CF-- is
going to equal CF over AD.

00:07:23.553 --> 00:07:28.120
AD is the same thing
as CD-- over CD.

00:07:28.120 --> 00:07:32.680
And so we know the ratio of AB
to AD is equal to CF over CD.

00:07:32.680 --> 00:07:34.790
But we just proved to
ourselves, because this

00:07:34.790 --> 00:07:38.495
is an isosceles triangle, that
CF is the same thing as BC

00:07:38.495 --> 00:07:39.470
right over here.

00:07:42.750 --> 00:07:43.760
And we're done.

00:07:43.760 --> 00:07:48.030
We've just proven AB over
AD is equal to BC over CD.

00:07:48.030 --> 00:07:51.460
So there's two things we
had to do here is one,

00:07:51.460 --> 00:07:54.930
construct this other
triangle, that, assuming

00:07:54.930 --> 00:07:58.220
this was parallel, that gave
us two things, that gave us

00:07:58.220 --> 00:08:01.010
another angle to show
that they're similar

00:08:01.010 --> 00:08:02.820
and also allowed us
to establish-- sorry,

00:08:02.820 --> 00:08:05.520
I have something
stuck in my throat.

00:08:05.520 --> 00:08:06.710
Just coughed off camera.

00:08:06.710 --> 00:08:10.630
So I should go get a
drink of water after this.

00:08:10.630 --> 00:08:13.090
So constructing
this triangle here,

00:08:13.090 --> 00:08:15.080
we were able to both
show it's similar

00:08:15.080 --> 00:08:17.660
and to construct this
larger isosceles triangle

00:08:17.660 --> 00:08:24.380
to show, look, if we can
find the ratio of this side

00:08:24.380 --> 00:08:26.670
to this side is the same
as a ratio of this side

00:08:26.670 --> 00:08:28.710
to this side, that's
analogous to showing

00:08:28.710 --> 00:08:31.510
that the ratio of this side
to this side is the same as BC

00:08:31.510 --> 00:08:32.429
to CD.

00:08:32.429 --> 00:08:34.709
And we are done.


Incenter and incircles of a triangle

WEBVTT
Kind: captions
Language: en

00:00:00.680 --> 00:00:02.246
I have triangle ABC here.

00:00:02.246 --> 00:00:03.620
And in the last
video, we started

00:00:03.620 --> 00:00:05.890
to explore some of the
properties of points

00:00:05.890 --> 00:00:07.600
that are on angle bisectors.

00:00:07.600 --> 00:00:09.266
And now, what I want
to do in this video

00:00:09.266 --> 00:00:11.730
is just see what happens when
we apply some of those ideas

00:00:11.730 --> 00:00:14.680
to triangles or the
angles in triangles.

00:00:14.680 --> 00:00:17.990
So let's bisect this angle
right over here-- angle BAC.

00:00:17.990 --> 00:00:20.710
And let me draw
an angle bisector.

00:00:20.710 --> 00:00:23.460
So the angle bisector
might look something--

00:00:23.460 --> 00:00:26.620
I want to make sure I get
that angle right in two.

00:00:26.620 --> 00:00:27.680
Pretty close.

00:00:27.680 --> 00:00:29.780
So that looks pretty close.

00:00:29.780 --> 00:00:31.080
So that's the angle bisector.

00:00:31.080 --> 00:00:32.920
Let me call this point
right over here--

00:00:32.920 --> 00:00:35.910
I don't know-- I could
call this point D.

00:00:35.910 --> 00:00:38.380
And then, let me draw
another angle bisector,

00:00:38.380 --> 00:00:41.790
the one that bisects angle ABC.

00:00:41.790 --> 00:00:43.360
So let me just draw this one.

00:00:43.360 --> 00:00:47.450
It might look something
like that right over there.

00:00:47.450 --> 00:00:49.510
And I could maybe
call this point E.

00:00:49.510 --> 00:00:58.290
So AD bisects angle BAC,
and BE bisects angle ABC.

00:00:58.290 --> 00:01:00.370
So the fact that
this green line-- AD

00:01:00.370 --> 00:01:02.390
bisects this angle
right over here--

00:01:02.390 --> 00:01:04.450
that tells us that
this angle must

00:01:04.450 --> 00:01:06.570
be equal to that angle
right over there.

00:01:06.570 --> 00:01:08.380
They must have
the same measures.

00:01:08.380 --> 00:01:13.830
And the fact that this bisects
this angle-- angle ABC-- tells

00:01:13.830 --> 00:01:17.180
us that the measure of
this angle-- angle ABE--

00:01:17.180 --> 00:01:19.670
must be equal to the
measure of angle EBC.

00:01:22.600 --> 00:01:25.220
Now, we see clearly that
they have intersected

00:01:25.220 --> 00:01:28.250
at a point inside of the
triangle right over there.

00:01:28.250 --> 00:01:32.350
So let's call that
point I just for fun.

00:01:32.350 --> 00:01:34.640
I'm skipping a few
letters, but it's

00:01:34.640 --> 00:01:36.410
a useful letter
based on what we are

00:01:36.410 --> 00:01:38.430
going to call this
in very short order.

00:01:38.430 --> 00:01:39.888
And there's some
interesting things

00:01:39.888 --> 00:01:43.430
that we know about I. I sits on
both of these angle bisectors.

00:01:43.430 --> 00:01:46.470
And we saw in the previous
video that any point that

00:01:46.470 --> 00:01:50.430
sits on an angle bisector
is equidistant from the two

00:01:50.430 --> 00:01:52.430
sides of that angle.

00:01:52.430 --> 00:01:54.920
So for example, I sits on AD.

00:01:54.920 --> 00:01:57.360
So it's going to be
equidistant from the two

00:01:57.360 --> 00:01:59.510
sides of angle BAC.

00:01:59.510 --> 00:02:02.140
So this is one side
right over here.

00:02:02.140 --> 00:02:04.000
This is one side
right over there.

00:02:04.000 --> 00:02:06.100
And then this is the other
side right over there.

00:02:06.100 --> 00:02:10.949
So because I sits on AD, we
know that these two distances

00:02:10.949 --> 00:02:13.040
are going to be the
same, assuming that this

00:02:13.040 --> 00:02:15.699
is the shortest distance
between I and the sides.

00:02:15.699 --> 00:02:17.740
And then, we've also shown
in that previous video

00:02:17.740 --> 00:02:20.341
that, when we talk about
the distance between a point

00:02:20.341 --> 00:02:22.840
and a line, we're talking about
the shortest distance, which

00:02:22.840 --> 00:02:25.185
is the distance you get if
you drop a perpendicular.

00:02:25.185 --> 00:02:27.060
So that's why I drew
the perpendiculars right

00:02:27.060 --> 00:02:27.930
over there.

00:02:27.930 --> 00:02:29.440
And let's label these.

00:02:29.440 --> 00:02:31.180
This could be
point F. This could

00:02:31.180 --> 00:02:33.410
be point G right over here.

00:02:33.410 --> 00:02:37.910
So because I sits on AD,
sits on this angle bisector,

00:02:37.910 --> 00:02:45.340
we know that IF is
going to be equal to IG.

00:02:45.340 --> 00:02:46.510
Fair enough.

00:02:46.510 --> 00:02:49.355
Now, I also sits on
this angle bisector.

00:02:49.355 --> 00:02:54.985
It also sits on BE, which says
that it must be equidistant.

00:02:54.985 --> 00:02:59.180
The distance to AB must be the
same as I's distance to BC.

00:02:59.180 --> 00:03:00.830
I's distance to
AB we already just

00:03:00.830 --> 00:03:02.380
said is this right over here.

00:03:02.380 --> 00:03:03.710
It's IG.

00:03:03.710 --> 00:03:05.750
But we also know
that that distance

00:03:05.750 --> 00:03:09.550
must be the same as the
distance between I and BC.

00:03:09.550 --> 00:03:13.657
So if I drop another
perpendicular right over here.

00:03:13.657 --> 00:03:15.490
And let's say I call
this point-- let's see,

00:03:15.490 --> 00:03:17.520
I haven't used H
right over here--

00:03:17.520 --> 00:03:19.860
this distance must be
the same as this distance

00:03:19.860 --> 00:03:22.760
because I sits on this bisector.

00:03:22.760 --> 00:03:25.210
So IG must be equal to IH.

00:03:29.690 --> 00:03:31.540
But IF is also equal to IG.

00:03:31.540 --> 00:03:34.600
So we can also say that IF--
I mean, if IF is equal to IG

00:03:34.600 --> 00:03:40.610
is equal to IH, we also
know that IF is equal to IH.

00:03:40.610 --> 00:03:41.911
Pretty much common sense.

00:03:41.911 --> 00:03:43.910
If this is equal to that,
that is equal to that,

00:03:43.910 --> 00:03:46.820
then these two have to
be equal to each other.

00:03:46.820 --> 00:03:49.644
But if I is equidistant
from two sides of an angle--

00:03:49.644 --> 00:03:51.060
this is the second
part of what we

00:03:51.060 --> 00:03:53.890
proved in the previous video--
if you have a point that

00:03:53.890 --> 00:03:57.230
is equidistant from
two sides of an angle,

00:03:57.230 --> 00:04:02.300
then that point must sit on the
angle bisector for that angle.

00:04:02.300 --> 00:04:06.150
So this right here
tells us that I

00:04:06.150 --> 00:04:20.250
must be on angle
bisector of angle ACB.

00:04:20.250 --> 00:04:25.680
Because it's equidistant to
those two sides of angle ACB.

00:04:25.680 --> 00:04:28.290
And what we have just
shown is that there's

00:04:28.290 --> 00:04:31.030
a unique point inside
the triangle that

00:04:31.030 --> 00:04:33.055
sits on all three
angle bisectors.

00:04:33.055 --> 00:04:35.680
It's not always obvious that if
you took three lines-- in fact,

00:04:35.680 --> 00:04:36.910
normally, if you
took three lines,

00:04:36.910 --> 00:04:38.743
they're not going to
intersect in one point.

00:04:38.743 --> 00:04:40.480
Two lines, a very
reasonable thing to do.

00:04:40.480 --> 00:04:42.850
But three lines, not always
going to intersect in one

00:04:42.850 --> 00:04:43.350
point.

00:04:43.350 --> 00:04:45.820
But once again-- like we
saw with the circumcenter

00:04:45.820 --> 00:04:48.725
where we took the perpendicular
bisectors of the side-- that

00:04:48.725 --> 00:04:50.850
was pretty neat that they
intersected in one point.

00:04:50.850 --> 00:04:52.500
Now, it's also cool
that we're showing

00:04:52.500 --> 00:04:56.340
that the angle bisectors all
intersect in one unique point.

00:04:56.340 --> 00:05:01.250
I is on the angle bisector of
ACB, so the bisector of ACB

00:05:01.250 --> 00:05:05.250
will look something like this.

00:05:05.250 --> 00:05:07.130
And this angle
right over here is

00:05:07.130 --> 00:05:10.930
going to be congruent to
this angle right over there.

00:05:10.930 --> 00:05:13.500
So we've just shown that
if you take the three angle

00:05:13.500 --> 00:05:18.430
bisectors of a triangle, it
will intersect in a unique point

00:05:18.430 --> 00:05:20.990
right over there that
sits on all three of them.

00:05:20.990 --> 00:05:23.470
So it seems worthwhile that
we should call this something

00:05:23.470 --> 00:05:24.030
special.

00:05:24.030 --> 00:05:24.900
And we do.

00:05:24.900 --> 00:05:26.740
And that's why I
called it I. We call

00:05:26.740 --> 00:05:33.700
I the incenter of triangle ABC.

00:05:36.252 --> 00:05:37.710
And you're going
to see in a second

00:05:37.710 --> 00:05:39.290
why it's called the incenter.

00:05:39.290 --> 00:05:41.020
When we talked about
the circumcenter,

00:05:41.020 --> 00:05:42.940
that was the center
of a circle that

00:05:42.940 --> 00:05:45.790
could be circumscribed
about the triangle.

00:05:45.790 --> 00:05:48.950
I-- we'll see in
about five seconds--

00:05:48.950 --> 00:05:52.760
is the center of
a circle that can

00:05:52.760 --> 00:05:54.570
be put inside the
triangle that's

00:05:54.570 --> 00:05:56.320
tangent to the three sides.

00:05:56.320 --> 00:05:57.930
And how do we construct that?

00:05:57.930 --> 00:06:00.000
Well, we've just
established that I

00:06:00.000 --> 00:06:02.780
is equidistant to
each of the sides--

00:06:02.780 --> 00:06:04.650
that this length is
equal to that length

00:06:04.650 --> 00:06:06.130
is equal to that length.

00:06:06.130 --> 00:06:08.050
So what happens if you
set up a circle with I

00:06:08.050 --> 00:06:11.560
as a center that has a radius
equal to the distance between I

00:06:11.560 --> 00:06:14.920
and any one of the
sides, which is equal,

00:06:14.920 --> 00:06:17.434
that has a radius
equal to IF, IG, or IH?

00:06:17.434 --> 00:06:19.350
Well, then, you're going
to have a circle that

00:06:19.350 --> 00:06:26.510
looks something like this.

00:06:26.510 --> 00:06:30.690
Let me draw it a little
bit better than that.

00:06:30.690 --> 00:06:31.760
Well, you can imagine.

00:06:31.760 --> 00:06:34.290
This is my best attempt
to draw a circle.

00:06:34.290 --> 00:06:36.270
This circle right here
that has the radius

00:06:36.270 --> 00:06:38.880
equal to the distance between
I and any of the sides--

00:06:38.880 --> 00:06:41.910
which we've already
established as being equal--

00:06:41.910 --> 00:06:44.330
we see that it's sitting
inside of the circle.

00:06:44.330 --> 00:06:47.100
So why don't we call
this an incircle?

00:06:47.100 --> 00:06:50.440
So circle I. Remember,
you label circles usually

00:06:50.440 --> 00:06:51.960
with the point at the center.

00:06:51.960 --> 00:07:01.090
Circle I is the incircle
of triangle ABC.

00:07:01.090 --> 00:07:04.030
And of course, the
radius of circle I--

00:07:04.030 --> 00:07:06.844
so we could call this length r.

00:07:06.844 --> 00:07:10.390
We say r is equal
to IF, which is

00:07:10.390 --> 00:07:15.170
equal to IH, which
is equal to IG.

00:07:15.170 --> 00:07:18.920
We can call that
length the inradius.

00:07:18.920 --> 00:07:21.510
And it makes sense
because it's inside.

00:07:21.510 --> 00:07:23.520
When we were talking
about the intersection

00:07:23.520 --> 00:07:27.267
of the perpendicular bisectors,
we had our circumcenter

00:07:27.267 --> 00:07:29.100
because that was the
center of a circle that

00:07:29.100 --> 00:07:31.960
is circumscribed
about the triangle.

00:07:31.960 --> 00:07:33.480
Now, we're taking
the intersection

00:07:33.480 --> 00:07:35.810
of the angle bisectors.

00:07:35.810 --> 00:07:38.600
And then, using that, we're
able to define a circle that

00:07:38.600 --> 00:07:42.120
is kind of within the
triangle and whose

00:07:42.120 --> 00:07:45.150
sides are tangent to the circle.

00:07:45.150 --> 00:07:47.660
And since it's inside it,
we call this an incircle.

00:07:47.660 --> 00:07:52.460
We call the intersection of the
angle bisectors the incenter.

00:07:52.460 --> 00:07:56.340
And we call this distance
right over here the inradius.


Point-line distance and angle bisectors

WEBVTT
Kind: captions
Language: en

00:00:00.000 --> 00:00:03.022
In this video I'm going to talk a little more about points on

00:00:03.022 --> 00:00:07.074
angle bisector but before that I want to at least make sure we

00:00:07.074 --> 00:00:09.840
understand what we mean when we talk about the distance

00:00:09.840 --> 00:00:12.923
between a point and a line.

00:00:12.923 --> 00:00:16.524
So say that that is some point, point A

00:00:16.524 --> 00:00:19.923
this is some line right over here, we'll call that line,

00:00:19.923 --> 00:00:23.239
BC, so when you're taking the distance between

00:00:23.239 --> 00:00:25.491
a point and another point, it's very obvious,

00:00:25.491 --> 00:00:28.291
you just draw a line to that other point,

00:00:28.291 --> 00:00:30.940
Well I already used B, you just draw a line to that other point

00:00:30.940 --> 00:00:33.307
and you find the length of that line,

00:00:33.307 --> 00:00:35.459
so distance seems very straight foreward between

00:00:35.459 --> 00:00:37.441
two points, but what about a point and a line

00:00:37.441 --> 00:00:40.041
because there are many points on this line

00:00:40.041 --> 00:00:45.358
maybe were going to find this distance, or maybe were going to find this distance, or this distance,

00:00:45.358 --> 00:00:48.091
and these are all going to be different lengths

00:00:48.091 --> 00:00:49.941
so how do we have one unique distance?

00:00:49.941 --> 00:00:52.657
And the way that we think about this, and we're going

00:00:52.657 --> 00:00:54.724
to do this in much more depth in future math courses

00:00:54.724 --> 00:00:57.408
especially when you start vectors and linear algebra

00:00:57.408 --> 00:01:00.139
and all the rest, is distance between a point and a line is

00:01:00.139 --> 00:01:03.524
really the shortest distance and that shortest distance

00:01:03.524 --> 00:01:05.091
is as if you were to drop a

00:01:05.091 --> 00:01:08.925
perpendicular from that point to the line, so this right over here,

00:01:08.925 --> 00:01:12.159
this right over here, is what we call the distance

00:01:12.159 --> 00:01:15.692
the distance between the point and the line, and this

00:01:15.692 --> 00:01:17.293
is perpendicular right over here, to

00:01:17.293 --> 00:01:20.676
recognize that this is indeed the shortest distance

00:01:20.676 --> 00:01:23.859
think about this relative -the distance between this point

00:01:23.859 --> 00:01:28.109
and any other point on this line. so pick another

00:01:28.109 --> 00:01:31.807
point on this line right over here, let's call this

00:01:31.807 --> 00:01:36.126
point E, and think about this --now this is an

00:01:36.126 --> 00:01:38.025
arbitrary point I could have drawn E here,

00:01:38.025 --> 00:01:39.710
I could have drawn E here, I could have drawn E anywhere,

00:01:39.710 --> 00:01:41.676
but regardless of where you draw E, if you

00:01:41.676 --> 00:01:44.697
draw a line segment between A and E, you see that

00:01:44.697 --> 00:01:47.710
we've form a right triangle from A to E, to the

00:01:47.710 --> 00:01:51.660
point we had the perpendicular, so let me call this

00:01:51.660 --> 00:01:54.044
point right here F, you're always going to draw

00:01:54.044 --> 00:01:57.659
a right triangle assuming that E is different than F

00:01:57.659 --> 00:02:00.210
and if you will immediately see that you see that D

00:02:00.210 --> 00:02:02.593
has to be shorter than this orange length, because this

00:02:02.593 --> 00:02:05.309
orange length is the hypotenuse. The hypotenuse

00:02:05.309 --> 00:02:07.943
is always going to be the longest side a triangle

00:02:07.943 --> 00:02:10.778
D squared plus whatever length this squared, is

00:02:10.778 --> 00:02:14.044
going to be equal to this length squared. So hopefully

00:02:14.044 --> 00:02:16.894
that at least gives you a decent sense why dropping

00:02:16.894 --> 00:02:20.728
the perpendicular will always give you the shortest

00:02:20.728 --> 00:02:22.978
distance between a point and a line, and that unique

00:02:22.978 --> 00:02:25.978
shortest distance is what we call the distance

00:02:25.978 --> 00:02:28.493
between a point and a line. Now with that out of the way,

00:02:28.493 --> 00:02:31.393
let's think a little bit about angle bisectors.

00:02:31.393 --> 00:02:38.328
So let me draw an angle here, so let me, draw an angle.

00:02:38.328 --> 00:02:42.462
So lets call this point, let me do it in a different color,

00:02:42.462 --> 00:02:44.759
let's call that point A, lets call this point B,

00:02:44.759 --> 00:02:47.662
and let's call this point C right over here.

00:02:47.662 --> 00:02:50.544
And an angle bisector is essentially line, or a segment,

00:02:50.544 --> 00:02:53.928
or a ray that splits an angle into two equal angles

00:02:53.928 --> 00:02:56.312
and we've talked about this before,

00:02:56.312 --> 00:02:59.545
so for example if we want to bisect angle ABC,

00:02:59.545 --> 00:03:01.761
so this angle right over here, we want to split it

00:03:01.761 --> 00:03:05.962
in two, we are going to -I can draw a better version

00:03:05.962 --> 00:03:09.244
of that, we want to split it in two, we want to split it in two

00:03:09.244 --> 00:03:13.394
like, let me draw a little bit better, my drawing

00:03:13.394 --> 00:03:17.846
still doesn't look like, that looks decent, alright,

00:03:17.846 --> 00:03:23.345
so lets call this point right over here D,

00:03:23.345 --> 00:03:25.844
and again we could even say that, that's a ray,

00:03:25.844 --> 00:03:27.728
or we could call that a segment, or whatever,

00:03:27.728 --> 00:03:29.830
but the way to think about this, is if now

00:03:29.830 --> 00:03:35.496
angle DBC is equal to angle DBA, so this angle,

00:03:35.496 --> 00:03:43.963
DBC is equal to angle DBA, we can say that DB bisects

00:03:43.963 --> 00:03:49.963
angle ABC, so we can say that DB and now Im talking

00:03:49.963 --> 00:03:52.562
about segment DB we could have made it a ray if we

00:03:52.562 --> 00:03:54.895
had -keep going to the right, or a line. DB

00:03:54.895 --> 00:04:02.880
bisects BISECTS angle ABC, ABC

00:04:02.880 --> 00:04:05.247
Fare enough, now, the whole reason why I started

00:04:05.247 --> 00:04:08.814
this video talking about distances between points and lines,

00:04:08.814 --> 00:04:10.614
is that I want to show you, that any point that is on

00:04:10.614 --> 00:04:12.780
an angle bisector, is actually going to be

00:04:12.780 --> 00:04:16.352
equidistant from the sides of the angle, and then were

00:04:16.352 --> 00:04:18.362
going to go the other way saying, any point

00:04:18.362 --> 00:04:20.998
that is equidistant from the sides of an angle is

00:04:20.998 --> 00:04:24.203
going on the angle bisector. So lets take an

00:04:24.203 --> 00:04:26.570
arbitrary point that sits on this angle bisector

00:04:26.570 --> 00:04:30.903
so let's take this point right over here,

00:04:30.903 --> 00:04:33.453
Ill call that arbitrary point F, or actually I havnt used

00:04:33.453 --> 00:04:36.686
E yet, so this is going to be and arbitrary point on our

00:04:36.686 --> 00:04:40.820
angle bisector, and now let's look at the distance between

00:04:40.820 --> 00:04:44.436
E and BC and the distance between E and BA we already

00:04:44.436 --> 00:04:46.922
say that, -we already said that the distance between

00:04:46.922 --> 00:04:49.388
point and a line is if you drop a perpendicular

00:04:49.388 --> 00:04:51.788
from a point to that line, which you can always do

00:04:51.788 --> 00:04:54.754
so lets draw a perpendicular right over here,

00:04:54.754 --> 00:04:57.904
this is one distance, and then this is the other distance

00:04:57.904 --> 00:05:04.238
this is, -this orange line right here is the distance between E and BC and this orange line is the distance

00:05:04.238 --> 00:05:11.721
between E and BA and what I want to prove is that these distances are equal

00:05:11.721 --> 00:05:17.372
well the first thing to realize, is that we have two right triangles over here, they both share

00:05:17.372 --> 00:05:24.239
th- they both have the same angle, they're not sharing it but angle ABE is congruent

00:05:24.239 --> 00:05:29.788
to angle CBE and we know that because DB bisects it so this angle is equal to that angle

00:05:29.788 --> 00:05:34.372
they're both right triangles so the actually have 2 angles in common, which actually means

00:05:34.372 --> 00:05:38.605
they actually have 3 angles in common and constrains what this other angle would be and they also have

00:05:38.605 --> 00:05:44.039
this side in common, -they have 3 angles are congruent to each other, they are not

00:05:44.039 --> 00:05:48.256
necessarily in common, but it does have this side in common, BE is the hypotenuse

00:05:48.256 --> 00:05:52.240
of both of these right triangles, and so, you can invoke

00:05:52.240 --> 00:05:56.990
using this angle, that angle and this side, this angle,that angle and this side

00:05:56.990 --> 00:06:00.554
you could say that these 2 triangles are going to be congruent to each other

00:06:00.554 --> 00:06:04.855
so we could say that triangle, and let me put some points over here, so let's see

00:06:04.855 --> 00:06:14.007
let's call this F and let's call this G, we can say that triangle EBF is congruent to triangle

00:06:14.007 --> 00:06:23.656
EBG and we can use AAS, by Angle Angle Side congruency or you could say hey if

00:06:23.656 --> 00:06:28.223
two angles, -if corresponding angles are the same then the third angle is also going to be the same

00:06:28.223 --> 00:06:31.925
so this angle right over here could also be the same and

00:06:31.925 --> 00:06:35.307
you could use ASA, but either way these two things are going to be congruent

00:06:35.307 --> 00:06:39.290
but if these two things are congruent, then the corresponding sides are going to be congruent

00:06:39.290 --> 00:06:47.774
so then, then length of EF, segment EF, is going to be congruent to segment

00:06:47.774 --> 00:06:51.191
EF is going to be congruent to segment EG which is the same

00:06:51.191 --> 00:06:57.224
thing as the length of EF is equal to the length of EG, these are really, these are really

00:06:57.224 --> 00:07:02.940
equivalent statements right over there, so the length of EF, the length of EF

00:07:02.940 --> 00:07:06.575
is equal to the length of EG and the length of those two segments are the distances

00:07:06.575 --> 00:07:09.893
between the point and those 2 respective sides.

00:07:09.893 --> 00:07:15.443
We've just proven the first case, if a point lies on an angle bisector, it is

00:07:15.443 --> 00:07:21.608
equidistant from the 2 sides of the angle, now let's go the other way around,

00:07:21.608 --> 00:07:29.025
let's say that I have, so let me draw another angle here, so let me draw another angle over here and let's call

00:07:29.025 --> 00:07:40.142
this A, B, and C and let's pick some arbitrary point E, let's point some arbitrary point E

00:07:40.142 --> 00:07:43.929
right over here and let's say we start off with the assumption that E is equidistant

00:07:43.929 --> 00:07:51.793
to BC and BA and what we want to do is prove that E must be on the angle bisector,

00:07:51.793 --> 00:07:55.959
so here, if you're on the angle bisector you're equidistant, over here we're going to show if you're

00:07:55.959 --> 00:08:02.059
equidistant you're on the angle bisector, so if it's equidistant to BC and BA, then this perpendicular

00:08:02.059 --> 00:08:06.524
right over here, this perpendicular right over here, is going to be congruent to this

00:08:06.524 --> 00:08:13.909
perpendicular right over there, and let me give these points' labels, so let call this point D, and let's

00:08:13.909 --> 00:08:20.909
call this point right over F, and let's just draw segment BE here, let's just draw segment

00:08:20.909 --> 00:08:29.577
BE right over here, so once again, we have 2 right triangles, we already know that

00:08:29.577 --> 00:08:32.511
2 of the legs are congruent to each other, they both share the hypotenuse this

00:08:32.511 --> 00:08:36.493
hypotenuse is equal to its self, we know from the pythagorean theorm

00:08:36.493 --> 00:08:43.210
if you know 2 sides of a triangle, it determines the third side, so and we know 2 sides of both

00:08:43.210 --> 00:08:48.694
of these, so the third sides must be the same, so this side must be equal to that side

00:08:48.694 --> 00:08:53.495
so you could invoke SSS, Side Side Side, to show these 2 triangles are congruent

00:08:53.495 --> 00:08:56.944
or you actually didn't need to -have to good there you could have used a special

00:08:56.944 --> 00:09:02.985
case, the RSH case, where if you have a right triangle, so if you have the right triangle

00:09:02.985 --> 00:09:06.012
you have one set of sides that are congruent, and you have the hypotenuse that is congruent

00:09:06.012 --> 00:09:11.419
then you're also okay, you could use RSH to prove congruency as well,

00:09:11.419 --> 00:09:21.742
and so either way we know that triangle EBD, triangle EBD, is congruent to triangle EBF

00:09:21.753 --> 00:09:29.135
congruent to triangle EBF, we used Side Side Side here but you could have used RSH, let me write that

00:09:29.135 --> 00:09:34.903
RSH, which is -we know that Angle Side Side can be used for any general triangle

00:09:34.903 --> 00:09:38.719
but it can be, -RSH is essentially Angle Side Side for right triangles, if you have

00:09:38.719 --> 00:09:49.669
2 sides of a right triangle and- in common, --if 2 sides of a right triangle are congruent then

00:09:49.669 --> 00:09:57.036
the 2 triangles are definitely congruent but once you know that two triangles are congruent, then

00:09:57.036 --> 00:10:01.970
they're corresponding angles have to be congruent, and angle EBD, angle EBD

00:10:01.970 --> 00:10:11.636
corresponds to angle EBF so we know that angle EBD must be congruent to angle EBF

00:10:11.636 --> 00:10:21.854
so EBD must be congruent to EBF, well if EBD is congruent to EBF then that means

00:10:21.854 --> 00:10:38.086
that segment EB must bisect, must bisect angle CBF, or actually CBA even, it could be called CEF

00:10:38.086 --> 00:10:43.804
angle CBA. So we're done! If over here, we show that if something sits on a bisector is

00:10:43.804 --> 00:10:48.021
equidistant from the sides of the angle, and here we showed if it's equidistant from the sides

00:10:48.021 --> 00:10:54.470
of the angle it sits on the angle bisector, or, or -it could even be the end point of an angle bisector

00:10:54.470 --> 00:10:56.815
but clearly, it sits on it.


Three points defining a circle

WEBVTT
Kind: captions
Language: en

00:00:00.500 --> 00:00:03.670
We know that three
points define a triangle.

00:00:03.670 --> 00:00:05.650
So if I were to take
three random points here,

00:00:05.650 --> 00:00:09.540
so let's call that point
A, point B, and then

00:00:09.540 --> 00:00:11.960
let's say this is point
C right over here.

00:00:11.960 --> 00:00:13.620
If we say that these
three points are

00:00:13.620 --> 00:00:16.940
the vertices of a triangle,
they define a unique triangle.

00:00:16.940 --> 00:00:19.680
So this would be
triangle A-- try

00:00:19.680 --> 00:00:22.486
to draw my lines as straight
as possible-- triangle ABC.

00:00:25.590 --> 00:00:28.230
Now we've also learned
in the last few videos

00:00:28.230 --> 00:00:33.400
that triangle ABC has
a unique circumcenter.

00:00:33.400 --> 00:00:36.700
And that is a point that is
equidistant to these three

00:00:36.700 --> 00:00:40.220
vertices, it equidistant
to these three points.

00:00:40.220 --> 00:00:42.950
So the way can we can find
it is we draw a perpendicular

00:00:42.950 --> 00:00:45.940
bisector of each of
these sides and where

00:00:45.940 --> 00:00:47.952
the three perpendicular
bisectors intersect--

00:00:47.952 --> 00:00:49.535
and we show that
they always intersect

00:00:49.535 --> 00:00:52.777
at a unique point-- that
is that circumcenter.

00:00:52.777 --> 00:00:54.610
And I'll do it really
quick right over here.

00:00:54.610 --> 00:00:56.620
So let's say that this
is the perpendicular

00:00:56.620 --> 00:01:01.400
bisector of that side, this
is the perpendicular bisector

00:01:01.400 --> 00:01:07.610
of that side, and this is
the perpendicular bisector

00:01:07.610 --> 00:01:08.780
of that side.

00:01:08.780 --> 00:01:12.380
So these are all perpendicular,
this is perpendicular,

00:01:12.380 --> 00:01:15.820
and they each bisect the sides.

00:01:15.820 --> 00:01:20.540
B to this point is going to
be equal to this point to A. A

00:01:20.540 --> 00:01:23.060
to this point is going
to be equal to that point

00:01:23.060 --> 00:01:29.410
to C. C to this point is going
to be equal to that point to B.

00:01:29.410 --> 00:01:32.250
And this point right over here,
we've already talked about,

00:01:32.250 --> 00:01:35.420
we'll call that point O. We
call that the circumcenter.

00:01:35.420 --> 00:01:39.260
O is the circumcenter.

00:01:39.260 --> 00:01:41.310
This is all a little
bit of review.

00:01:41.310 --> 00:01:43.970
So if you have three points,
you have a unique triangle.

00:01:43.970 --> 00:01:47.290
That unique triangle has a
unique circumcenter, which

00:01:47.290 --> 00:01:49.940
is equidistant to the three
points of the triangle,

00:01:49.940 --> 00:01:53.090
three-- I should say the three
vertices of the triangle--

00:01:53.090 --> 00:01:55.850
and that distance
between the circumcenter

00:01:55.850 --> 00:01:59.490
and the three points, the
three vertices, I should say.

00:01:59.490 --> 00:02:02.160
So let me draw that
in a different color.

00:02:02.160 --> 00:02:09.000
So this distance, OA, the
length of OA, the length of OC,

00:02:09.000 --> 00:02:14.070
and the length of OB,
so OA is equal to OC

00:02:14.070 --> 00:02:17.050
is equal to OB, which
is the c circumradius.

00:02:21.010 --> 00:02:24.040
And we've learned when we
first talked about circles,

00:02:24.040 --> 00:02:26.040
if you give me a
point, and if we

00:02:26.040 --> 00:02:27.830
find the locus of
all points that

00:02:27.830 --> 00:02:31.599
are equidistant from that
point, then that is a circle.

00:02:31.599 --> 00:02:34.140
And when I say a locus, all I
mean is, the set of all points.

00:02:34.140 --> 00:02:37.690
If you give me any
point right over here,

00:02:37.690 --> 00:02:41.000
so that's an arbitrary point,
and you also specify a radius,

00:02:41.000 --> 00:02:44.100
and say what is the set of
all the points on this two

00:02:44.100 --> 00:02:46.820
dimensional plane
that are equidistant,

00:02:46.820 --> 00:02:48.990
that are that radius
away from the center?

00:02:48.990 --> 00:02:50.500
It uniquely defines a circle.

00:02:50.500 --> 00:02:53.530
That's how we defined a
circle right over here.

00:02:53.530 --> 00:02:55.400
And similarly, if
you say, look, if you

00:02:55.400 --> 00:02:56.970
start with the
center at O, and you

00:02:56.970 --> 00:03:01.460
say all of the points that are
the circumradius away from O,

00:03:01.460 --> 00:03:03.720
it will uniquely
identify a circle.

00:03:03.720 --> 00:03:07.320
And that circle will contain
the points A, B, and C

00:03:07.320 --> 00:03:11.250
because those are the
circumradius away from O.

00:03:11.250 --> 00:03:13.730
So they are included
in that set.

00:03:13.730 --> 00:03:15.290
So the circle would
look something

00:03:15.290 --> 00:03:18.030
like-- let me draw it.

00:03:18.030 --> 00:03:20.250
It would look
something like this--

00:03:20.250 --> 00:03:25.660
trying my best to draw
it, just like that.

00:03:25.660 --> 00:03:27.180
Everything we've
talked about, just

00:03:27.180 --> 00:03:29.920
now within the last few
minutes, is all review.

00:03:29.920 --> 00:03:31.020
We know all of this.

00:03:31.020 --> 00:03:33.810
But I went over it just
to kind of reinstate

00:03:33.810 --> 00:03:40.320
a pretty interesting idea, that
if you give me three points

00:03:40.320 --> 00:03:45.220
that defines a unique
triangle, and if you

00:03:45.220 --> 00:03:47.840
have a unique triangle--
And let me make it clear.

00:03:47.840 --> 00:03:52.170
This is three
non-collinear points,

00:03:52.170 --> 00:03:58.250
so three points not
on the same line.

00:03:58.250 --> 00:04:00.760
If you have three points that
are not on the same line,

00:04:00.760 --> 00:04:02.850
that defines a unique triangle.

00:04:02.850 --> 00:04:12.480
For any unique triangle you
have a unique circumcenter

00:04:12.480 --> 00:04:15.642
and circumradius.

00:04:15.642 --> 00:04:17.850
I'll rewrite it, I don't
want to get lazy and confuse

00:04:17.850 --> 00:04:18.860
you-- circumradius.

00:04:21.480 --> 00:04:23.180
And if you give me
any point in space,

00:04:23.180 --> 00:04:26.940
any unique point, and a radius,
the set of all points that

00:04:26.940 --> 00:04:28.870
are exactly that
radius away from it,

00:04:28.870 --> 00:04:30.400
that defines a unique circle.

00:04:35.765 --> 00:04:37.390
So we went through
all of this business

00:04:37.390 --> 00:04:38.931
of talking about
the unique triangle,

00:04:38.931 --> 00:04:41.400
and the unique circumcenter,
and the unique radius,

00:04:41.400 --> 00:04:44.310
to really just show you that
if you give me any three

00:04:44.310 --> 00:04:51.210
points that eventually, really,
just defines a unique circle.

00:04:51.210 --> 00:04:54.070
So just as you need three
points to define a triangle,

00:04:54.070 --> 00:04:56.890
you also need three
points to define a circle,

00:04:56.890 --> 00:04:58.360
two points won't do it.

00:04:58.360 --> 00:05:02.980
And one way to think about it
is, if you give me two points,

00:05:02.980 --> 00:05:04.615
there's an infinite
number of triangles

00:05:04.615 --> 00:05:06.390
that I construct with
those two points,

00:05:06.390 --> 00:05:08.264
because I could put the
third point anywhere.

00:05:08.264 --> 00:05:10.030
I could construct this triangle.

00:05:10.030 --> 00:05:11.590
I could construct
this triangle, I

00:05:11.590 --> 00:05:15.320
could construct this triangle,
I can construct this triangle.

00:05:15.320 --> 00:05:16.820
And all of these
triangles are going

00:05:16.820 --> 00:05:19.470
to have different circumcenters
and different radiuses.

00:05:19.470 --> 00:05:22.680
And so they're going to
have different circles that

00:05:22.680 --> 00:05:24.870
circumscribe about
those triangles.

00:05:24.870 --> 00:05:27.450
So this one-- so
for example, this

00:05:27.450 --> 00:05:29.830
would be one circle
that could go around,

00:05:29.830 --> 00:05:31.960
that could circumscribe
that triangle.

00:05:31.960 --> 00:05:34.660
You could have this
circle right over here.

00:05:34.660 --> 00:05:36.110
So you see clearly,
very clearly,

00:05:36.110 --> 00:05:38.330
that two points are not enough.

00:05:38.330 --> 00:05:41.390
You need three points, three
points lead to a triangle,

00:05:41.390 --> 00:05:43.150
lead to a unique circle.

00:05:43.150 --> 00:05:45.170
So that by itself
is kind of cool.

00:05:45.170 --> 00:05:48.530
Now, another question is,
if I have just a circle,

00:05:48.530 --> 00:05:55.230
and if it's circumscribed
about an arbitrary triangle,

00:05:55.230 --> 00:05:59.345
is the center of that circle
necessarily the circumcenter?

00:05:59.345 --> 00:06:00.970
So let's think about
that a little bit,

00:06:00.970 --> 00:06:08.280
because there are some
non-intuitive cases here.

00:06:08.280 --> 00:06:13.100
So if I draw a circle
right over here,

00:06:13.100 --> 00:06:15.370
its center is right over there.

00:06:15.370 --> 00:06:17.840
And if I draw an
arbitrary triangle where

00:06:17.840 --> 00:06:21.760
all of the vertices of that
triangle are on this circle,

00:06:21.760 --> 00:06:26.070
is this center necessarily the
circumcenter of that triangle?

00:06:26.070 --> 00:06:28.480
So let me draw a
crazy situation.

00:06:28.480 --> 00:06:30.800
So let me draw one where
this thing is clearly

00:06:30.800 --> 00:06:33.130
outside of the
triangle, so that we

00:06:33.130 --> 00:06:35.370
could have a triangle
that looks like this.

00:06:35.370 --> 00:06:38.264
And it's clearly all three
vertices sit on the circle.

00:06:38.264 --> 00:06:39.930
So you might at first
say, wait, there's

00:06:39.930 --> 00:06:41.530
no way this could
be the circumcenter,

00:06:41.530 --> 00:06:43.170
it's not even
inside the triangle.

00:06:43.170 --> 00:06:46.170
But remember, this
point right here

00:06:46.170 --> 00:06:48.704
is equidistant to every
point on the circle.

00:06:48.704 --> 00:06:50.370
I should say, every
point on this circle

00:06:50.370 --> 00:06:53.950
is equidistant from this point,
they're all the radius away.

00:06:53.950 --> 00:06:58.050
And all three points of this
triangle are on the circle,

00:06:58.050 --> 00:07:00.750
so they are all exactly a radius
away from this point right

00:07:00.750 --> 00:07:01.320
over here.

00:07:01.320 --> 00:07:04.040
So this distance right over
here is going to be a radius,

00:07:04.040 --> 00:07:06.900
this distance right over
here is going to be a radius,

00:07:06.900 --> 00:07:10.690
and this distance right over
here is going to be a radius.

00:07:10.690 --> 00:07:14.170
Now, this point is clearly
equidistant from that point

00:07:14.170 --> 00:07:15.620
and that point.

00:07:15.620 --> 00:07:17.330
We know that, it's
exactly R away

00:07:17.330 --> 00:07:20.527
from both of those
vertices of the triangle.

00:07:20.527 --> 00:07:22.110
So if it's equidistant--
and we proved

00:07:22.110 --> 00:07:23.526
this in a previous
video-- if it's

00:07:23.526 --> 00:07:25.420
equidistant from
both of those points,

00:07:25.420 --> 00:07:28.840
it must be on the
perpendicular bisector

00:07:28.840 --> 00:07:31.620
of the segment that
joins those two points.

00:07:31.620 --> 00:07:35.400
So this must be on the
perpendicular bisector--

00:07:35.400 --> 00:07:38.630
so that's perpendicular
and it bisects

00:07:38.630 --> 00:07:40.047
that segment right over there.

00:07:40.047 --> 00:07:42.380
But we can make the same
argument for this segment right

00:07:42.380 --> 00:07:46.560
over here, because this point
is R from the center-- we'll

00:07:46.560 --> 00:07:49.670
call it O, I am tired of
just saying this point.

00:07:49.670 --> 00:07:52.890
Point O is equidistant
from-- let me label these,

00:07:52.890 --> 00:07:55.880
so let's call this A, B, C.

00:07:55.880 --> 00:07:59.310
So we already said point O
is equidistant from C and B,

00:07:59.310 --> 00:08:02.680
so it must be on the
perpendicular bisector of BC.

00:08:02.680 --> 00:08:05.340
And it's also
equidistant from A and B.

00:08:05.340 --> 00:08:08.390
It's R away from both, because
A and B both sit on the circle,

00:08:08.390 --> 00:08:10.750
they're both a radius
away from the center.

00:08:10.750 --> 00:08:13.440
So it also must sit on the
perpendicular bisector of AB.

00:08:19.620 --> 00:08:22.160
Let me draw it a little
bit neater, there you go.

00:08:22.160 --> 00:08:25.990
So it must also be on this
perpendicular bisector.

00:08:25.990 --> 00:08:29.350
And then finally, it also
is equidistant from A.

00:08:29.350 --> 00:08:32.790
It's for the same distance
from A is it is from C.

00:08:32.790 --> 00:08:35.760
Because those are both R away,
they both sit on the circle,

00:08:35.760 --> 00:08:39.010
so it must be on the
perpendicular bisector of AC

00:08:39.010 --> 00:08:41.980
as well.

00:08:41.980 --> 00:08:43.704
So AC is right over here.

00:08:43.704 --> 00:08:45.120
This is what the
interesting thing

00:08:45.120 --> 00:08:50.480
is, we're seeing that the
three perpendicular bisectors

00:08:50.480 --> 00:08:52.030
of the three sides
of this triangle,

00:08:52.030 --> 00:08:53.940
they do definitely
intersect, but they

00:08:53.940 --> 00:08:57.460
are intersecting at a point
outside of that triangle.

00:08:57.460 --> 00:09:00.210
And that point is the
center of the circle.

00:09:00.210 --> 00:09:06.080
So once again, the last idea is,
O is equidistant from A and C,

00:09:06.080 --> 00:09:09.230
so it must sit on the
perpendicular bisector of AC,

00:09:09.230 --> 00:09:11.680
which would look
something like this, which

00:09:11.680 --> 00:09:15.830
would look something like that.

00:09:15.830 --> 00:09:18.380
So once again, we see the three
perpendicular bisectors are

00:09:18.380 --> 00:09:21.090
intersecting at a unique
point, and O really

00:09:21.090 --> 00:09:23.580
is the circumcenter.

00:09:23.580 --> 00:09:26.535
So if you take any circle,
if you take a circle,

00:09:26.535 --> 00:09:28.950
and if you put any
triangle whose vertices

00:09:28.950 --> 00:09:31.630
sit on the circle, the
center of that circle

00:09:31.630 --> 00:09:33.270
is its circumcenter.

00:09:33.270 --> 00:09:35.740
So we just drew
a situation where

00:09:35.740 --> 00:09:38.940
this is the circumcenter that
sits outside of the triangle

00:09:38.940 --> 00:09:39.820
proper.

00:09:39.820 --> 00:09:46.010
So point O is also going
to be the circumcenter

00:09:46.010 --> 00:09:48.160
of this triangle
right over here.

00:09:48.160 --> 00:09:51.090
And point O is also going
to be the circumcenter

00:09:51.090 --> 00:09:54.710
of this triangle
right over here.

00:09:54.710 --> 00:09:57.030
It's going to sit on
all three perpendicular

00:09:57.030 --> 00:09:59.390
bisectors of this, and
we know that because it's

00:09:59.390 --> 00:10:02.840
equidistant from all
three points of any

00:10:02.840 --> 00:10:06.480
of these triangles where the
vertices sit on the circle

00:10:06.480 --> 00:10:07.536
itself.


Circumcenter of a right triangle

WEBVTT
Kind: captions
Language: en

00:00:00.170 --> 00:00:02.710
What I want to do in this video is prove

00:00:02.720 --> 00:00:06.080
that the circumcenter of a right triangle,

00:00:06.090 --> 00:00:09.990
is actually the midpoint of the hypotenuse, and to do that,

00:00:10.000 --> 00:00:13.660
I'm gonna take, first take a look at the perpendicular bisector

00:00:13.670 --> 00:00:16.880
of one of the legs, of this, of this right triangle

00:00:16.890 --> 00:00:19.600
So, let me construct the perpendicular bisector

00:00:19.610 --> 00:00:23.330
of leg BC right over here, so it's going to look something like this,

00:00:23.340 --> 00:00:27.000
it's going to look something like this, this,

00:00:27.010 --> 00:00:30.480
it intersects at a right angle, its perpendicular, and it bisects it

00:00:30.490 --> 00:00:33.540
So B, this is from B to this point

00:00:33.550 --> 00:00:35.800
which we'll call M, maybe M for midpoint,

00:00:35.810 --> 00:00:38.770
is the same, as it is from M to C,

00:00:38.780 --> 00:00:41.230
so those two distances are going to be equal,

00:00:41.240 --> 00:00:45.130
and let's call the point where this perpendicular bisector

00:00:45.140 --> 00:00:46.660
intersects the hypotenuse,

00:00:46.670 --> 00:00:49.110
let's call this O,and we're gonna prove that O

00:00:49.120 --> 00:00:53.300
is the circumcenter of this right triangle

00:00:53.310 --> 00:00:55.550
Now, the first thing that you might realize,

00:00:55.560 --> 00:00:57.530
and this is what we've seen in many problems,

00:00:57.540 --> 00:01:02.670
the triangle OBM, looks similar to triangle ABC,

00:01:02.680 --> 00:01:04.400
and its actually not too hard to prove,

00:01:04.470 --> 00:01:06.600
they both already have a 90 degree angle,

00:01:06.610 --> 00:01:09.260
so if we show that they, they both have another angle

00:01:09.270 --> 00:01:10.870
Another set of corresponding angles

00:01:10.880 --> 00:01:13.520
that are congruent to each other, then we know that they're similar

00:01:13.530 --> 00:01:17.490
By AA similarity, and they both clearly share this angle,

00:01:17.500 --> 00:01:21.250
right over here, OBC is part of the smaller triangle,

00:01:21.260 --> 00:01:23.710
and ABC which is really the same angle,

00:01:23.720 --> 00:01:27.070
is part of the larger triangle, and so, and they also obviously share

00:01:27.080 --> 00:01:30.870
a 90 degree angle, so by AA triangle similarity,

00:01:30.880 --> 00:01:40.700
we have triangle OBM, OBM is similar, is similar to triangle ABC,

00:01:40.710 --> 00:01:47.350
is similar to triangle ABC, and what's useful about this?

00:01:47.360 --> 00:01:50.530
Is we know similar triangles are ratios between corresponding sides

00:01:50.540 --> 00:01:55.270
are constant, so for example, we know that the ratio between side BM,

00:01:55.280 --> 00:01:59.290
which is on the smaller triangle, we know that the ratio between BM

00:01:59.300 --> 00:02:02.740
Let me do this in a different color, just to, just for the sake of it,

00:02:02.750 --> 00:02:04.840
we know that the ratio between BM

00:02:04.850 --> 00:02:10.500
and BC, BM and BC, the ratio of this side on the smaller

00:02:10.510 --> 00:02:12.740
triangle to the corresponding side on the larger triangle

00:02:12.750 --> 00:02:16.030
Is going to be the same as the ratio of the hypotenuse

00:02:16.040 --> 00:02:20.510
on the smaller triangle, BO to the hypotenuse of the larger triangle,

00:02:20.520 --> 00:02:25.280
because they are similar, well we know what the ratio of BM to BC is,

00:02:25.290 --> 00:02:29.150
BM is half of BC, so this ratio over here

00:02:29.160 --> 00:02:33.550
is going to be equal to one half, this is M is the bi,

00:02:33.560 --> 00:02:35.590
of the midpoint of these things,

00:02:35.600 --> 00:02:38.200
so this is exactly the same distance as this,

00:02:38.210 --> 00:02:40.640
so this is one half of the entire BC,

00:02:40.650 --> 00:02:45.430
so if one half is equal to BM, over BC is equal to BO over BA

00:02:45.440 --> 00:02:48.740
We then know, if we just kind of ignore this middle part,

00:02:48.750 --> 00:02:53.970
right over here, that one half is equal to BO over BA,

00:02:53.980 --> 00:02:57.900
over BA, if you cross multiply it,

00:02:57.910 --> 00:03:01.590
if you cross multiply, you see that, well there's multiple ways

00:03:01.600 --> 00:03:04.760
to think about, but you could just cross multiply, and you say BA,

00:03:04.770 --> 00:03:10.130
is equal to 2BO, or if you divide both sides by two,

00:03:10.140 --> 00:03:14.990
and their really equivalent statements one half BA is equal to BO,

00:03:15.000 --> 00:03:20.420
so BO is one half of BA, so this is one half BA,

00:03:20.430 --> 00:03:23.530
and so this other length, AO right over here

00:03:23.540 --> 00:03:26.760
This is going to be B, this is going to be, this going to be BA,

00:03:26.770 --> 00:03:30.810
minus one half BA, so this is also going to be, one half BA,

00:03:30.820 --> 00:03:32.950
and so, this segment right over here,

00:03:32.960 --> 00:03:37.420
AO, AO is going to be congruent to OB

00:03:37.430 --> 00:03:40.110
So what we just shown, first of all,

00:03:40.120 --> 00:03:42.590
is that this perpendicular bisector, right over here,

00:03:42.600 --> 00:03:45.360
The perpendicular bisector of segment BC,

00:03:45.370 --> 00:03:49.890
it intersects the hypotenuse of our right triangle at the midpoint,

00:03:49.900 --> 00:03:51.440
So we've already established,

00:03:51.450 --> 00:03:53.350
so we, one thing that we've already established,

00:03:53.360 --> 00:03:57.950
is O, is the midpoint, is the midpoint,

00:03:57.960 --> 00:04:03.950
of the hypotenuse, of the hypotenuse, of the hypotenuse AB,

00:04:03.960 --> 00:04:05.780
well, that by itself is interesting,

00:04:05.790 --> 00:04:11.180
but, we also know that if a point sits on a perpendicular bisector

00:04:11.190 --> 00:04:14.220
of a segment, is equidistant, it's equal distant

00:04:14.230 --> 00:04:17.490
from the end point of the segment, we'd show that in a previous video

00:04:17.500 --> 00:04:23.730
So we also know that O, OB that's equidistant to the end points

00:04:23.740 --> 00:04:28.550
of the segment, right over here, that OB is equal to OC, but we know,

00:04:28.560 --> 00:04:30.190
from this first statement right over here,

00:04:30.200 --> 00:04:35.510
that OB is alsoequal to OA, OB is also equal to OA,

00:04:35.520 --> 00:04:37.410
its of OB is equal to OC

00:04:37.420 --> 00:04:40.900
OB is equal to OA, that means OC must be equal to OA,

00:04:40.910 --> 00:04:43.850
OC must be equal to OA

00:04:43.860 --> 00:04:46.600
Or another way to think about it, is at this point O,

00:04:46.610 --> 00:04:50.570
Is equal distant from all of the points on our tri,

00:04:50.580 --> 00:04:53.860
all of the vertices, I should say, this point O is equidistant,

00:04:53.870 --> 00:04:58.320
from all of the vertices of our triangle, of our triangle,

00:04:58.330 --> 00:05:02.170
So this distance, this distance, which is really going to become

00:05:02.180 --> 00:05:05.290
our circumradius, is the same as this distance right over here,

00:05:05.300 --> 00:05:09.540
which is the same as this distance right over there

00:05:09.550 --> 00:05:12.850
So that we know that O is equidistant,

00:05:12.860 --> 00:05:19.480
equidistant to all, all vertices, which is another way of saying

00:05:19.490 --> 00:05:25.570
that O is the circumcenter, O is the circumcenter, so we've just

00:05:25.580 --> 00:05:29.300
proven that if you have the circumcenter of a right triangle,

00:05:29.310 --> 00:05:32.660
it is the midpoint of the hypotenuse of the right triangle,

00:05:32.670 --> 00:05:34.920
or the other way around, that the hypotenuse of the right triangle

00:05:34.930 --> 00:05:39.320
is the circumcenter, because you only have one circumcenter of any,

00:05:39.330 --> 00:05:40.880
of any triangle


Circumcenter of a triangle

WEBVTT
Kind: captions
Language: en

00:00:00.500 --> 00:00:02.440
Let's start off with segment AB.

00:00:02.440 --> 00:00:06.920
So that's point A. This is
point B right over here.

00:00:06.920 --> 00:00:10.820
And let's set up a perpendicular
bisector of this segment.

00:00:10.820 --> 00:00:14.330
So it will be both perpendicular
and it will split the segment

00:00:14.330 --> 00:00:15.410
in two.

00:00:15.410 --> 00:00:17.700
So thus we could
call that line l.

00:00:17.700 --> 00:00:21.840
That's going to be a
perpendicular bisector,

00:00:21.840 --> 00:00:24.610
so it's going to intersect
at a 90-degree angle,

00:00:24.610 --> 00:00:25.750
and it bisects it.

00:00:25.750 --> 00:00:28.020
This length and this
length are equal,

00:00:28.020 --> 00:00:30.650
and let's call this
point right over here

00:00:30.650 --> 00:00:33.260
M, maybe M for midpoint.

00:00:33.260 --> 00:00:35.220
What I want to prove
first in this video

00:00:35.220 --> 00:00:38.200
is that if we pick an arbitrary
point on this line that

00:00:38.200 --> 00:00:43.220
is a perpendicular bisector of
AB, then that arbitrary point

00:00:43.220 --> 00:00:45.860
will be an equal distant
from A, or that distance

00:00:45.860 --> 00:00:47.450
from that point to
A will be the same

00:00:47.450 --> 00:00:53.160
as that distance
from that point to B.

00:00:53.160 --> 00:00:55.550
So let me pick an arbitrary
point on this perpendicular

00:00:55.550 --> 00:00:56.690
bisector.

00:00:56.690 --> 00:01:00.100
So let's call that
arbitrary point C.

00:01:00.100 --> 00:01:02.360
And so you can imagine we
like to draw a triangle,

00:01:02.360 --> 00:01:05.750
so let's draw a triangle where
we draw a line from C to A

00:01:05.750 --> 00:01:08.140
and then another
one from C to B.

00:01:08.140 --> 00:01:11.550
And essentially, if we can
prove that CA is equal to CB,

00:01:11.550 --> 00:01:13.220
then we've proven
what we want to prove,

00:01:13.220 --> 00:01:17.166
that C is an equal distance
from A as it is from B. Well,

00:01:17.166 --> 00:01:19.290
there's a couple of
interesting things we see here.

00:01:19.290 --> 00:01:22.180
We know that AM is
equal to MB, and we also

00:01:22.180 --> 00:01:25.010
know that CM is equal to itself.

00:01:25.010 --> 00:01:28.410
Obviously, any segment is
going to be equal to itself.

00:01:28.410 --> 00:01:30.560
And we know if this
is a right angle,

00:01:30.560 --> 00:01:31.780
this is also a right angle.

00:01:31.780 --> 00:01:35.560
This line is a perpendicular
bisector of AB.

00:01:35.560 --> 00:01:37.400
And so we have two
right triangles.

00:01:37.400 --> 00:01:38.430
And actually, we don't
even have to worry about

00:01:38.430 --> 00:01:39.650
that they're right triangles.

00:01:39.650 --> 00:01:43.060
If you look at triangle
AMC, you have this side

00:01:43.060 --> 00:01:46.020
is congruent to the
corresponding side on triangle

00:01:46.020 --> 00:01:47.420
BMC.

00:01:47.420 --> 00:01:50.120
Then you have an angle in
between that corresponds

00:01:50.120 --> 00:01:55.790
to this angle over here, angle
AMC corresponds to angle BMC,

00:01:55.790 --> 00:01:58.030
and they're both 90 degrees,
so they're congruent.

00:01:58.030 --> 00:02:01.170
And then you have the side
MC that's on both triangles,

00:02:01.170 --> 00:02:02.720
and those are congruent.

00:02:02.720 --> 00:02:06.100
So we can just use SAS,
side-angle-side congruency.

00:02:08.639 --> 00:02:15.910
So we can write
that triangle AMC

00:02:15.910 --> 00:02:25.755
is congruent to triangle BMC
by side-angle-side congruency.

00:02:29.690 --> 00:02:31.840
And so if they are
congruent, then

00:02:31.840 --> 00:02:34.210
all of their corresponding
sides are congruent

00:02:34.210 --> 00:02:36.840
and AC corresponds to BC.

00:02:36.840 --> 00:02:38.970
So these two things
must be congruent.

00:02:38.970 --> 00:02:42.340
This length must be the same as
this length right over there,

00:02:42.340 --> 00:02:44.390
and so we've proven
what we want to prove.

00:02:44.390 --> 00:02:47.350
This arbitrary point C that
sits on the perpendicular

00:02:47.350 --> 00:02:52.510
bisector of AB is equidistant
from both A and B.

00:02:52.510 --> 00:02:55.660
And I could have known that if
I drew my C over here or here,

00:02:55.660 --> 00:02:58.290
I would have made the exact
same argument, so any C that

00:02:58.290 --> 00:02:59.674
sits on this line.

00:02:59.674 --> 00:03:00.590
So that's fair enough.

00:03:00.590 --> 00:03:01.590
So let me just write it.

00:03:01.590 --> 00:03:07.092
So this means that
AC is equal to BC.

00:03:07.092 --> 00:03:08.550
Now, let's go the
other way around.

00:03:08.550 --> 00:03:10.580
Let's say that we
find some point that

00:03:10.580 --> 00:03:14.320
is equidistant
from A and B. Let's

00:03:14.320 --> 00:03:17.210
prove that it has to sit on
the perpendicular bisector.

00:03:17.210 --> 00:03:19.640
So let's do this again.

00:03:19.640 --> 00:03:21.394
So I'll draw it like this.

00:03:21.394 --> 00:03:27.410
So this is my A. This is my B,
and let's throw out some point.

00:03:27.410 --> 00:03:29.140
We'll call it C again.

00:03:29.140 --> 00:03:30.900
So let's say that
C right over here,

00:03:30.900 --> 00:03:34.450
and maybe I'll draw
a C right down here.

00:03:34.450 --> 00:03:37.870
So this is C, and we're going
to start with the assumption

00:03:37.870 --> 00:03:40.890
that C is equidistant
from A and B.

00:03:40.890 --> 00:03:44.399
So CA is going to
be equal to CB.

00:03:44.399 --> 00:03:46.190
This is what we're
going to start off with.

00:03:46.190 --> 00:03:48.290
This is going to
be our assumption,

00:03:48.290 --> 00:03:50.680
and what we want
to prove is that C

00:03:50.680 --> 00:03:59.550
sits on the perpendicular
bisector of AB.

00:03:59.550 --> 00:04:02.110
So we've drawn a triangle here,
and we've done this before.

00:04:02.110 --> 00:04:04.680
We can always drop an
altitude from this side

00:04:04.680 --> 00:04:06.540
of the triangle right over here.

00:04:06.540 --> 00:04:09.930
So we can set up a
line right over here.

00:04:09.930 --> 00:04:12.180
Let me draw it like this.

00:04:12.180 --> 00:04:14.887
So let's just drop an
altitude right over here.

00:04:14.887 --> 00:04:16.470
Although we're really
not dropping it.

00:04:16.470 --> 00:04:18.600
We're kind of lifting an
altitude in this case.

00:04:18.600 --> 00:04:20.025
But if you rotated
this around so

00:04:20.025 --> 00:04:28.290
that the triangle looked like
this, so this was B, this is A,

00:04:28.290 --> 00:04:31.460
and that C was up
here, you would really

00:04:31.460 --> 00:04:33.380
be dropping this altitude.

00:04:33.380 --> 00:04:36.260
And so you can
construct this line

00:04:36.260 --> 00:04:38.710
so it is at a right
angle with AB,

00:04:38.710 --> 00:04:42.700
and let me call this the point
at which it intersects M. So

00:04:42.700 --> 00:04:46.140
to prove that C lies on
the perpendicular bisector,

00:04:46.140 --> 00:04:47.980
we really have to
show that CM is

00:04:47.980 --> 00:04:50.480
a segment on the
perpendicular bisector,

00:04:50.480 --> 00:04:52.450
and the way we've
constructed it,

00:04:52.450 --> 00:04:53.970
it is already perpendicular.

00:04:53.970 --> 00:04:58.230
We really just have to
show that it bisects AB.

00:04:58.230 --> 00:05:00.762
So what we have right over
here, we have two right angles.

00:05:00.762 --> 00:05:02.720
If this is a right angle
here, this one clearly

00:05:02.720 --> 00:05:04.755
has to be the way
we constructed it.

00:05:04.755 --> 00:05:06.310
It's at a right angle.

00:05:06.310 --> 00:05:16.030
And then we know that the CM
is going to be equal to itself.

00:05:16.030 --> 00:05:18.110
And so this is a right angle.

00:05:18.110 --> 00:05:20.190
We have a leg, and
we have a hypotenuse.

00:05:20.190 --> 00:05:27.850
We know by the RSH postulate,
we have a right angle.

00:05:27.850 --> 00:05:30.044
We have one
corresponding leg that's

00:05:30.044 --> 00:05:31.710
congruent to the other
corresponding leg

00:05:31.710 --> 00:05:32.960
on the other triangle.

00:05:32.960 --> 00:05:34.700
We have a hypotenuse
that's congruent

00:05:34.700 --> 00:05:36.940
to the other hypotenuse,
so that means

00:05:36.940 --> 00:05:38.590
that our two triangles
are congruent.

00:05:38.590 --> 00:05:48.350
So triangle ACM is congruent
to triangle BCM by the RSH

00:05:48.350 --> 00:05:49.260
postulate.

00:05:49.260 --> 00:05:51.690
Well, if they're congruent,
then their corresponding sides

00:05:51.690 --> 00:05:52.773
are going to be congruent.

00:05:55.980 --> 00:06:01.660
So that tells us that AM must
be equal to BM because they're

00:06:01.660 --> 00:06:02.829
their corresponding sides.

00:06:02.829 --> 00:06:04.370
So this side right
over here is going

00:06:04.370 --> 00:06:05.910
to be congruent to that side.

00:06:05.910 --> 00:06:08.620
So this really is bisecting AB.

00:06:08.620 --> 00:06:12.525
So this line MC really is on
the perpendicular bisector.

00:06:17.205 --> 00:06:19.330
And the whole reason why
we're doing this is now we

00:06:19.330 --> 00:06:21.810
can do some interesting things
with perpendicular bisectors

00:06:21.810 --> 00:06:24.090
and points that are
equidistant from points

00:06:24.090 --> 00:06:26.640
and do them with triangles.

00:06:26.640 --> 00:06:29.460
So just to review, we
found, hey if any point

00:06:29.460 --> 00:06:31.810
sits on a perpendicular
bisector of a segment,

00:06:31.810 --> 00:06:33.852
it's equidistant from the
endpoints of a segment,

00:06:33.852 --> 00:06:34.934
and we went the other way.

00:06:34.934 --> 00:06:37.680
If any point is equidistant
from the endpoints of a segment,

00:06:37.680 --> 00:06:41.430
it sits on the perpendicular
bisector of that segment.

00:06:41.430 --> 00:06:44.975
So let's apply those
ideas to a triangle now.

00:06:44.975 --> 00:06:48.040
So let me draw myself
an arbitrary triangle.

00:06:48.040 --> 00:06:50.380
I'll try to draw
it fairly large.

00:06:50.380 --> 00:06:53.260
So let's say that's a
triangle of some kind.

00:06:53.260 --> 00:06:56.035
Let me give ourselves some
labels to this triangle.

00:06:56.035 --> 00:06:59.590
That's point A,
point B, and point C.

00:06:59.590 --> 00:07:02.330
You could call
this triangle ABC.

00:07:02.330 --> 00:07:05.510
Now, let me just construct
the perpendicular bisector

00:07:05.510 --> 00:07:08.010
of segment AB.

00:07:08.010 --> 00:07:10.520
So it's going to bisect it.

00:07:10.520 --> 00:07:13.030
So this distance is going to
be equal to this distance,

00:07:13.030 --> 00:07:15.020
and it's going to
be perpendicular.

00:07:15.020 --> 00:07:17.280
So it looks something like that.

00:07:17.280 --> 00:07:19.416
And it will be perpendicular.

00:07:19.416 --> 00:07:21.290
Actually, let me draw
this a little different

00:07:21.290 --> 00:07:24.020
because of the way I've
drawn this triangle,

00:07:24.020 --> 00:07:26.534
it's making us get close
to a special case, which

00:07:26.534 --> 00:07:28.450
we will actually talk
about in the next video.

00:07:28.450 --> 00:07:30.575
Let me draw this triangle
a little bit differently.

00:07:39.320 --> 00:07:39.820
OK.

00:07:39.820 --> 00:07:41.925
This one might be a
little bit better.

00:07:41.925 --> 00:07:44.050
And we'll see what special
case I was referring to.

00:07:44.050 --> 00:07:46.460
So this is going
to be A. This is

00:07:46.460 --> 00:07:49.606
going to be B. This
is going to be C.

00:07:49.606 --> 00:07:50.980
Now, let me take
this point right

00:07:50.980 --> 00:07:53.200
over here, which is
the midpoint of A and B

00:07:53.200 --> 00:07:56.950
and draw the
perpendicular bisector.

00:07:56.950 --> 00:08:01.915
So the perpendicular bisector
might look something like that.

00:08:01.915 --> 00:08:04.290
And I don't want it to make
it necessarily intersect in C

00:08:04.290 --> 00:08:06.820
because that's not necessarily
going to be the case.

00:08:06.820 --> 00:08:09.300
But this is going to
be a 90-degree angle,

00:08:09.300 --> 00:08:11.500
and this length is
equal to that length.

00:08:11.500 --> 00:08:15.800
And let me do the same thing
for segment AC right over here.

00:08:15.800 --> 00:08:18.480
Let me take its midpoint, which
if I just roughly draw it,

00:08:18.480 --> 00:08:20.540
it looks like it's
right over there.

00:08:20.540 --> 00:08:23.180
And then let me draw its
perpendicular bisector,

00:08:23.180 --> 00:08:25.085
so it would look
something like this.

00:08:29.020 --> 00:08:32.570
So this length right over
here is equal to that length,

00:08:32.570 --> 00:08:37.080
and we see that they
intersect at some point.

00:08:37.080 --> 00:08:40.360
Just for fun, let's
call that point O.

00:08:40.360 --> 00:08:43.039
And now there's some interesting
properties of point O.

00:08:43.039 --> 00:08:47.370
We know that since O sits on
AB's perpendicular bisector,

00:08:47.370 --> 00:08:50.590
we know that the
distance from O to B

00:08:50.590 --> 00:08:52.595
is going to be the same
as the distance from O

00:08:52.595 --> 00:08:54.880
to A. That's what we proved
in this first little proof

00:08:54.880 --> 00:08:55.410
over here.

00:08:55.410 --> 00:09:01.270
So we know that OA is
going to be equal to OB.

00:09:01.270 --> 00:09:02.580
Well, that's kind of neat.

00:09:02.580 --> 00:09:04.770
But we also know that
because of the intersection

00:09:04.770 --> 00:09:06.590
of this green
perpendicular bisector

00:09:06.590 --> 00:09:08.619
and this yellow
perpendicular bisector,

00:09:08.619 --> 00:09:10.660
we also know because it
sits on the perpendicular

00:09:10.660 --> 00:09:15.310
bisector of AC that
it's equidistant from A

00:09:15.310 --> 00:09:21.240
as it is to C. So we know
that OA is equal to OC.

00:09:21.240 --> 00:09:22.710
Now, this is interesting.

00:09:22.710 --> 00:09:23.915
OA is equal to OB.

00:09:23.915 --> 00:09:27.070
OA is also equal
to OC, so OC and OB

00:09:27.070 --> 00:09:28.780
have to be the
same thing as well.

00:09:28.780 --> 00:09:34.960
So we also know that
OC must be equal to OB.

00:09:34.960 --> 00:09:38.270
OC must be equal to OB.

00:09:38.270 --> 00:09:43.550
Well, if a point is equidistant
from two other points that

00:09:43.550 --> 00:09:46.470
sit on either end of a
segment, then that point

00:09:46.470 --> 00:09:49.670
must sit on the perpendicular
bisector of that segment.

00:09:49.670 --> 00:09:52.480
That's that second proof
that we did right over here.

00:09:52.480 --> 00:09:57.860
So it must sit on the
perpendicular bisector of BC.

00:09:57.860 --> 00:10:01.830
So if I draw the perpendicular
bisector right over there,

00:10:01.830 --> 00:10:09.670
then this definitely lies on
BC's perpendicular bisector.

00:10:09.670 --> 00:10:11.660
And what's neat about
this simple little proof

00:10:11.660 --> 00:10:13.800
that we've set up
in this video is

00:10:13.800 --> 00:10:18.070
we've shown that there's a
unique point in this triangle

00:10:18.070 --> 00:10:21.870
that is equidistant from all
of the vertices of the triangle

00:10:21.870 --> 00:10:25.001
and it sits on the perpendicular
bisectors of the three sides.

00:10:25.001 --> 00:10:26.500
Or another way to
think of it, we've

00:10:26.500 --> 00:10:29.680
shown that the perpendicular
bisectors, or the three sides,

00:10:29.680 --> 00:10:32.680
intersect at a
unique point that is

00:10:32.680 --> 00:10:35.270
equidistant from the vertices.

00:10:35.270 --> 00:10:38.240
And this unique point on a
triangle has a special name.

00:10:38.240 --> 00:10:39.960
We call O a circumcenter.

00:10:47.210 --> 00:10:50.770
And because O is
equidistant to the vertices,

00:10:50.770 --> 00:10:53.224
so this distance-- let
me do this in a color

00:10:53.224 --> 00:10:54.140
I haven't used before.

00:10:57.030 --> 00:11:00.240
This distance right over here
is equal to that distance right

00:11:00.240 --> 00:11:03.400
over there is equal to
that distance over there.

00:11:03.400 --> 00:11:07.510
If we construct a circle
that has a center at O

00:11:07.510 --> 00:11:09.770
and whose radius is
this orange distance,

00:11:09.770 --> 00:11:12.310
whose radius is any of
these distances over here,

00:11:12.310 --> 00:11:14.250
we'll have a circle
that goes through all

00:11:14.250 --> 00:11:19.620
of the vertices of our
triangle centered at O.

00:11:19.620 --> 00:11:21.380
So our circle would
look something

00:11:21.380 --> 00:11:24.121
like this, my best
attempt to draw it.

00:11:24.121 --> 00:11:26.120
And so what we've constructed
right here is one,

00:11:26.120 --> 00:11:28.460
we've shown that we can
construct something like this,

00:11:28.460 --> 00:11:33.434
but we call this
thing a circumcircle,

00:11:33.434 --> 00:11:35.850
and this distance right here,
we call it the circumradius.

00:11:39.219 --> 00:11:41.010
And once again, we know
we can construct it

00:11:41.010 --> 00:11:44.640
because there's a point here,
and it is centered at O.

00:11:44.640 --> 00:11:49.660
Now this circle, because
it goes through all

00:11:49.660 --> 00:11:51.390
of the vertices of
our triangle, we

00:11:51.390 --> 00:11:58.550
say that it is circumscribed
about the triangle.

00:11:58.550 --> 00:12:03.300
So we can say right over
here that the circumcircle O,

00:12:03.300 --> 00:12:12.560
so circle O right over
here is circumscribed

00:12:12.560 --> 00:12:18.740
about triangle ABC, which
just means that all three

00:12:18.740 --> 00:12:24.330
vertices lie on this circle
and that every point is

00:12:24.330 --> 00:12:28.295
the circumradius away
from this circumcenter.