The isosceles right triangle $ABC$ has right angle at $C$ and area $12.5$. The rays trisecting $\angle ACB$ intersect $AB$ at $D$ and $E$. What is the area of $\bigtriangleup CDE$? $\textbf{(A) }\dfrac{5\sqrt{2}}{3}\qquad$ $\textbf{(B) }\dfrac{50\sqrt{3}-75}{4}\qquad$ $\textbf{(C) }\dfrac{15\sqrt{3}}{8}\qquad$ $\textbf{(D) }\dfrac{50-25\sqrt{3}}{2}\qquad$ $\textbf{(E) }\dfrac{25}{6}$
Solution
The trisectors make $30^\circ$ angles, so in $\triangle ACD$, $\angle ACD=30^\circ$, $\angle A=45^\circ$ and $\angle ADC=105^\circ$. By the law of sines, $AD:\sin30^\circ = AC:\sin 105^\circ$.
By symmetry $EB=AD$, so $DE=\left(1-2\cdot \frac{\sqrt3-1}2\right)AB=(2-\sqrt3)AB$. Triangles $CDE$ and $CAB$ have the same height from $C$, so $[CDE]=(2-\sqrt3)[ABC]=\frac{(2-\sqrt3)\cdot 25}2$.
Triangle $ABC$ is a $30^\circ$-$60^\circ$-$90^\circ$ triangle with right angle at $C$. The rays from $C$ that trisect $\angle ACB$ meet $\overline{AB}$ at $D$ and $E$. What is the ratio of the area of $\triangle CDE$ to the area of $\triangle ABC$?
