17) Triangle $\bigtriangleup ABC$ and $\bigtriangleup ABD$ are isosceles with $AB=AC=BD$, and BD intersects AC at E(point E is on segment AC and segment BD). If $BD \perp AC$, then $\angle C + \angle D$ is
A)$115^{\circ}$ B)$120^{\circ}$ C)$130^{\circ}$ D)$135^{\circ}$ E) not uniquely determined
Above is a diagram. I tried setting one angle equal to $x^{\circ}$, then angle chasing. For example:
If I let $\angle C=x^{\circ}$, then $\angle CBE=90-x^{\circ}$. Since $\angle ABC=\angle C$, $\angle ABE=2x-180^{\circ}$ and $\angle BAC=180-2x^{\circ}$.
Also, $\angle ABD=360-2x^{\circ}$ so $\angle BAD=\angle ADB=x-90^{\circ}$.
What I get is that $\angle C+\angle D=2x-90$, and then I'm stuck. Any help would be appreciated.
P.S. Does translating BD up so that B is on A help?
New diagram (old one had error):
Best Answer
From isosceles $\triangle ABC$ we have $$2\angle C+\angle A=180^\circ$$ From isosceles $\triangle ABD$ and right-angled $\triangle AED$ we have $$\angle D+(\angle D-\angle A)=90^\circ$$ Adding those two equalities, we obtain $2\angle C+2\angle D=270^\circ$, so $\angle C+\angle D=135^\circ$.