# In equilateral triangle,One vertex of a square is at the midpoint of the side, and the two adjacent vertices are on the other two sides of triangle

In the equilateral triangle $ABC,AB=12.$One vertex of a square is at the midpoint of the side $BC$, and the two adjacent vertices are on the other two sides of the triangle.Find the length of the side of the square.

Let $DEFG$ be the square.Let $D$ be the midpoint of the side $BC.BD=DC=6.$

Let $E$ be on the side $AC$ and $G$ be on the side $AB$ such that $AG=12-a,BG=a$ and $AE=12-b,EC=b$.

In triangle $DEC,$ applying Cosine law,

$\cos 60^\circ=\frac{1}{2}=\frac{b^2+36-DE^2}{12b}........(1)$

In triangle $DGB,$ applying Cosine law,

$\cos 60^\circ=\frac{1}{2}=\frac{a^2+36-DG^2}{12a}........(2)$

I am stuck here.

• You must have $BG=EC$, so side length is just $\frac{6\sin 60}{\sin 75}\approx 5.38$. – almagest Apr 27 '16 at 14:04

1) $\triangle ADB: \angle D=90^{\circ}, DB=6, AD=\sqrt{12^2-6^2}=\sqrt{108}, AB=12$

2) $DG -$ bisector $\angle ADB$

$$DG=\sqrt2 \frac{AD \cdot DB}{AD+DB}=\sqrt2 \frac{6\sqrt3 \cdot 6}{6\sqrt3+6}=\frac{6\sqrt6}{\sqrt3+1}=3\sqrt6(\sqrt3-1)$$

• You introduced a spurious $\sqrt6$ right at the end. – almagest Apr 27 '16 at 14:06
• @Roman83 the question seems to require that E and F be on the triangle sides – G Cab Apr 27 '16 at 14:09
• @almagest: Thank you! – Roman83 Apr 27 '16 at 14:19
• I want to ask one thing ,how did you know $AD$ is perpendicular to $BC$ and why $F$ came on $AD$?@Roman83 – Vinod Kumar Punia Apr 27 '16 at 14:21
• @GCab No. The two vertices adjacent to the vertex at $D$. Roman83's triangle is rotated! – almagest Apr 27 '16 at 14:21

Let $x$ be length of side of square.Triangles DBG and CDE are congruent because they have an angle and two corresponding sides which are equal.Implies BG=EC.In which case Angle BDG = $45^0$ and Angle BGD = $75^0$.Applying sine rule in Triangle BGD $$\frac{x}{\sin{60^0}}=\frac{6}{\sin{75^0}}$$ I hope this was helpful.

• A neat solution. – almagest Apr 27 '16 at 14:23
• By which similarity criteria,you proved triangles $DBG$ and $CDE$ similar.Because there are only three similarity criterias,$SSS,AAA,SAS$.@VarunKumar – Vinod Kumar Punia Apr 27 '16 at 14:41
• @Vinod Kumar Punia he appears to have used ASS congruence, which only exists in the case of right angles (as RHS). I think it Is a mistake.. – N.S.JOHN Apr 27 '16 at 15:19