smc

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 $△AEB$, $△CBE$, and $△DEA$ 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 $△ABC$, we get $\angle B=\angle C$, so: $2\angle C+\angle D+\angle A=270^{\circ }$ From isosceles $△ABD$, we get $\angle A=\angle D$, so: $2\angle C+2\angle D=270^{\circ }$ $\angle C+\angle D=135^{\circ }$, which is answer $\boxed{135^{\circ }}$