Answer: D

Small Hint:

Use the $30^\circ$ trisector and subtract two congruent corner triangles

Big Hint:

If $AC=BC=5$, find the height of $\triangle ACD$

Solution:

Since $\triangle ABC$ is isosceles right with area $12.5$, its legs have length $5$. The trisectors make $\angle ACD=30^\circ$ and $\angle BCE=30^\circ$, so $\triangle ACD$ and $\triangle BCE$ have equal area.

Drop a perpendicular from $D$ to $AC,$ with foot $F.$ Since $D$ lies on $AB$ and $\angle A=45^\circ,$ $\triangle AFD$ is isosceles right. Let $AF=DF=h.$ Then $CF=5-h,$ and the $30^\circ$ angle gives $$\frac{CF}{DF}=\sqrt{3}.$$ Thus $5-h=h\sqrt{3}$, so $h=\frac{5}{1+\sqrt{3}}=\frac{5\sqrt{3}-5}{2}$.

Therefore $$\begin{aligned} &[ACD]=\frac12\cdot 5\cdot h \\ &=\frac{25\sqrt{3}-25}{4}. \end{aligned}$$ Subtracting the two congruent corner triangles from $\triangle ABC$, $$\begin{aligned} &[CDE]=\frac{25}{2} \\ &\quad {}-2\cdot\frac{25\sqrt{3}-25}{4} \\ &=\frac{50-25\sqrt{3}}{2}. \end{aligned}$$

Thus, D is the correct answer.