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

Question

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.

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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.

Answer 1

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

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

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

Answer 2

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.