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 $\bigtriangleup CDE$?
$\textbf{(A) }\dfrac{5\sqrt{2}}{3}\qquad\textbf{(B) }\dfrac{50\sqrt{3}-75}{4}\qquad\textbf{(C) }\dfrac{15\sqrt{3}}{8}\qquad\textbf{(D) }\dfrac{50-25\sqrt{3}}{2}\qquad\textbf{(E) }\dfrac{25}{6}$
[2015AMC10AProblem19Picture.png](https://artofproblemsolving.com/wiki/index.php/File:2015AMC10AProblem19Picture.png)
$\triangle ADC$ can be split into a $45-45-90$ right triangle and a $30-60-90$ right triangle by dropping a perpendicular from $D$ to side $AC$. Let $F$ be where that perpendicular intersects $AC$.
Because the side lengths of a $45-45-90$ right triangle are in ratio $a:a:a\sqrt{2}$, $DF = AF$.
Because the side lengths of a $30-60-90$ right triangle are in ratio $a:a\sqrt{3}:2a$ and $AF + FC = 5$, $DF = \frac{5 - AF}{\sqrt{3}}$.
Setting the two equations for $DF$ equal to each other, $AF = \frac{5 - AF}{\sqrt{3}}$.
Solving gives $AF = DF = \frac{5\sqrt{3} - 5}{2}$.
The area of $\triangle ADC = \frac12 \cdot DF \cdot AC = \frac{25\sqrt{3} - 25}{4}$.
$\triangle ADC$ is congruent to $\triangle BEC$, so their areas are equal.
A triangle's area can be written as the sum of the figures that make it up, so $[ABC] = [ADC] + [BEC] + [CDE]$.
$\frac{25}{2} = \frac{25\sqrt{3} - 25}{4} + \frac{25\sqrt{3} - 25}{4} + [CDE]$.
Solving gives $[CDE] = \frac{50 - 25\sqrt{3}}{2}$, so the answer is $\boxed{\textbf{(D) }\frac{50 - 25\sqrt{3}}{2}}$

Note
Another way to get $DF$ is that you assume $AF=DF$ to be equal to $a$, as previously mentioned, and $CF$ is equal to $a\sqrt{3}$. $AF+DF=5=a+a\sqrt{3}$