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

$\text{(A)}\ 115^\circ\qquad\text{(B)}\ 120^\circ\qquad\text{(C)}\ 130^\circ\qquad\text{(D)}\ 135^\circ\qquad\text{(E)}\ \text{not uniquely determined}$
Redraw the figure as a concave pentagon $ADECB$:

The angles of the pentagon will still sum to $180^\circ \cdot 3 = 540^\circ$, regardless of whether the pentagon is concave or not.  As a quick proof, note that the nine angles of three original triangles $\triangle  AEB$, $\triangle CBE$, and $\triangle DEA$ all make up the angles of the pentagon without overlap.
Since reflex $\angle E = 270^\circ$, we have:
$\angle C + \angle B + \angle A + \angle D = 540^\circ - 270^\circ = 270^\circ$. 
From isosceles $\triangle ABC$, we get $\angle B = \angle C$, so:
$2\angle C + \angle D + \angle A = 270^\circ$
From isosceles $\triangle ABD$, we get $\angle A = \angle D$, so:
$2\angle C + 2\angle D = 270^\circ$
$\angle C + \angle D = 135^\circ$, which is answer $\boxed{D}$