19. In an isosceles right $\triangle ABC$, it is known that $BC = AC$, $\angle C = 90^{\circ}$, and the area is $12.5$. The trisectors of $\angle ACB$ intersect the hypotenuse $AB$ at points $D$ and $E$. Then the area of $\triangle CDE$ is ( ).
(A) $\frac{5 \sqrt{2}}{3}$
(B) $\frac{50 \sqrt{3} - 75}{4}$
(C) $\frac{15 \sqrt{3}}{8}$
(D) $\frac{50 - 25 \sqrt{3}}{2}$
(E) $\frac{25}{6}$
19. D.

From the problem, we know $A B=A C=5$.
As shown in Figure 5, draw $D F \perp A C$ at point $F$.
Since $\angle A C B=90^{\circ}$, and $C D, C E$ are the trisectors of $\angle A C B$, therefore, $\angle A C D=30^{\circ}$.
$$
\begin{array}{l}
\text { Then } A F=D F, C F=\sqrt{3} D F, \\
A C=(1+\sqrt{3}) D F=5, \\
D F=\frac{5}{\sqrt{3}+1}=\frac{5(\sqrt{3}-1)}{2} . \\
\text { Thus, } S_{\triangle A D C}=\frac{25(\sqrt{3}-1)}{4} . \\
\text { Therefore, } S_{\triangle C D E}=\frac{25}{2}-\frac{25(\sqrt{3}-1)}{2}=\frac{50-25 \sqrt{3}}{2} .
\end{array}
$$