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 $(\quad)$.
(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.

Given $A B=A C=5$.
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, $\angle A C D=30^{\circ}$.
Then $A F=D F$, $C F=\sqrt{3} D F$,
$A C=(1+\sqrt{3}) D F=5$,
$D F=\frac{5}{\sqrt{3}+1}=\frac{5(\sqrt{3}-1)}{2}$.
Thus, $S_{\triangle A D C}=\frac{25(\sqrt{3}-1)}{4}$.
Hence, $S_{\triangle C D E}=\frac{25}{2}-\frac{25(\sqrt{3}-1)}{2}=\frac{50-25 \sqrt{3}}{2}$.