## Pages

### Pascal's Theorem

In a previous post, we were introduced to Pascal's Hexagrammum Mysticum Theorem - a magical theorem - which states that if we draw a hexagon inscribed in a conic section then the three pairs of opposite sides of the hexagon intersect at three points which lie on a straight line.

For example, as in the following figure we have a hexagon inscribed in a circle and the intersection points of the three pairs of the opposite sides of the hexagon $\{12, 45\}$, $\{23, 56\}$, $\{34, 61\}$ are collinear.

There is a useful tool to prove the collinearity of points - the Menelaus' Theorem - which states as follows:

Menelaus' Theorem: Given a triangle $ABC$ and three points $A'$, $B'$, $C'$ lying on the three lines $BC$, $CA$, $AB$, respectively. Then the three points $A'$, $B'$, $C'$ are collinear if and only if $$\frac{\vec{A'B}}{\vec{A'C}} \times \frac{\vec{B'C}}{\vec{B'A}} \times \frac{\vec{C'A}}{\vec{C'B}} = 1.$$

Today, we will use Menelaus' theorem to prove Pascal's theorem for the circle case.

When we want to use Menelaus' theorem to prove three certain points lying on a same straight line, we need to specify a triangle such that these three points belong to the three sides of the triangle.

In the following figure, for the Pascal's hexagon $P_1 P_2 P_3 P_4 P_5 P_6$, to show that the three intersection points $M_1$, $M_2$, $M_3$ are collinear, we need to find a triangle whose three sides contain $M_1$, $M_2$, $M_3$. Since $M_1$, $M_2$, $M_3$ lie on the sides of the hexagon, there are two natural choices for the triangle: either $ABC$, or $XYZ$.

Suppose we choose the triangle $ABC$, then we can see that the point $M_1$ lying on the line $BC$, the point $M_2$ lying on the line $CA$ and the point $M_3$ lying on the line $AB$. To use Menelaus' theorem, we need to show $$\frac{\vec{M_1 B}}{\vec{M_1 C}} \times \frac{\vec{M_2 C}}{\vec{M_2 A}} \times \frac{\vec{M_3 A}}{\vec{M_3 B}} = 1.$$ To calculate these quotients, we will apply Menelaus' theorem for the following triples of collinear points: $$\{M_1, P_5, P_6\}, ~~\{M_2, P_3, P_4\}, ~~\{M_3, P_1, P_2\}.$$
Let us now write down the proof in details.

A proof of Pascal's theorem

We will apply Menelaus' theorem for the triangle $ABC$.

Since $M_1$, $M_2$, $M_3$ lie on the three sides of the triangle $ABC$, to prove that they are collinear, we need to show $$\frac{\vec{M_1 B}}{\vec{M_1 C}} \times \frac{\vec{M_2 C}}{\vec{M_2 A}} \times \frac{\vec{M_3 A}}{\vec{M_3 B}} = 1.$$
Indeed, apply Menelaus' theorem for the triangle $ABC$ with the following triples of collinear points $$\{M_1, P_6, P_5\}, ~~\{M_2, P_4, P_3\}, ~~\{M_3, P_2, P_1\},$$ we have $$\frac{\vec{M_1 B}}{\vec{M_1 C}} \times \frac{\vec{P_6 C}}{\vec{P_6 A}} \times \frac{\vec{P_5 A}}{\vec{P_5 B}} = \frac{\vec{M_2 C}}{\vec{M_2 A}} \times \frac{\vec{P_4 A}}{\vec{P_4 B}} \times \frac{\vec{P_3 B}}{\vec{P_3 C}} = \frac{\vec{M_3 A}}{\vec{M_3 B}} \times \frac{\vec{P_2 B}}{\vec{P_2 C}} \times \frac{\vec{P_1 C}}{\vec{P_1 A}} = 1.$$ Thus, $$\frac{\vec{M_1 B}}{\vec{M_1 C}} = \frac{\vec{P_6 A}}{\vec{P_6 C}} \times \frac{\vec{P_5 B}}{\vec{P_5 A}}, ~~~\frac{\vec{M_2 C}}{\vec{M_2 A}} = \frac{\vec{P_4 B}}{\vec{P_4 A}} \times \frac{\vec{P_3 C}}{\vec{P_3 B}}, ~~~\frac{\vec{M_3 A}}{\vec{M_3 B}} = \frac{\vec{P_2 C}}{\vec{P_2 B}} \times \frac{\vec{P_1 A}}{\vec{P_1 C}}.$$ It follows that $$\frac{\vec{M_1 B}}{\vec{M_1 C}} \times \frac{\vec{M_2 C}}{\vec{M_2 A}} \times \frac{\vec{M_3 A}}{\vec{M_3 B}} = \frac{\vec{P_6 A}}{\vec{P_6 C}} \times \frac{\vec{P_5 B}}{\vec{P_5 A}} \times \frac{\vec{P_4 B}}{\vec{P_4 A}} \times \frac{\vec{P_3 C}}{\vec{P_3 B}} \times \frac{\vec{P_2 C}}{\vec{P_2 B}} \times \frac{\vec{P_1 A}}{\vec{P_1 C}}$$ $$= \frac{\vec{A P_1} ~\vec{A P_6}}{\vec{A P_4} ~\vec{A P_5}} \times \frac{\vec{B P_4} ~\vec{B P_5}}{\vec{B P_2} ~\vec{B P_3}} \times \frac{\vec{C P_2} ~\vec{C P_3}}{\vec{C P_1} ~\vec{C P_6}} = 1.$$
 Power of a point: $\vec{A P_1} ~\vec{A P_6} = \vec{A P_4} ~\vec{A P_5}$; $\vec{B P_4} ~\vec{B P_5} = \vec{B P_2} ~\vec{B P_3}$; $\vec{C P_2} ~\vec{C P_3} = \vec{C P_1} ~\vec{C P_6}$.

The last equality is from the power of a point property when we apply the power of the points $A$, $B$, $C$ to the circumcircle of the hexagon. Thus, we have completed the proof of Pascal's theorem.

Today, we have shown how to use Menelaus' theorem effectively to prove the Pascal's "magical" theorem. Indeed, Menelaus' theorem is a very useful tool often employed to prove the collinearity of points. Have a try now to see if we can use Menelaus' theorem to prove the Pappus' theorem!
 Pappus' Theorem

Hope to see you again in the next post.

Homework.

1. Prove Pascal's theorem by applying Menelaus' theorem for the triangle $XYZ$.

2. Prove Pappus' theorem.