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.

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

Thus, D is the correct answer.