The measure of angle $ABC$ is $50^\circ$, $\overline{AD}$ bisects angle $BAC$, and $\overline{DC}$ bisects angle $BCA$.  The measure of angle $ADC$ is

$\text{(A)}\ 90^\circ \qquad \text{(B)}\ 100^\circ \qquad \text{(C)}\ 115^\circ \qquad \text{(D)}\ 122.5^\circ \qquad \text{(E)}\ 125^\circ$
Let $\angle CAD = \angle BAD = x$, and let $\angle ACD = \angle BCD = y$
From $\triangle ABC$, we know that $50 + 2x + 2y = 180$, leading to $x + y = 65$.
From $\triangle ADC$, we know that $x + y + \angle D = 180$.  Plugging in $x + y = 65$, we get $\angle D = 180 - 65 = 115$, which is answer $\boxed{C}$.