AHSME 1996 Problem 21

Triangles $ABC$ and $ABD$ are isosceles with $AB=AC=BD$, and $BD$ intersects $AC$ at $E$. If $BD$ is perpendicular to $AC$, then $\angle C+\angle D$ is

$\text{(A)}\ 115^\circ \text{(B)}\ 120^\circ \text{(C)}\ 130^\circ \text{(D)}\ 135^\circ \text{(E)}\ \text{not uniquely determined}$