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.
Best Answer
1) $\triangle ADB: \angle D=90^{\circ}, DB=6, AD=\sqrt{12^2-6^2}=\sqrt{108}, AB=12$
2) $DG - $ bisector $\angle ADB$
$$DG=\sqrt2 \frac{AD \cdot DB}{AD+DB}=\sqrt2 \frac{6\sqrt3 \cdot 6}{6\sqrt3+6}=\frac{6\sqrt6}{\sqrt3+1}=3\sqrt6(\sqrt3-1)$$