# Isosceles triangles + angle chasing

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):

• I can see that it's $\angle ABE+90^{\circ}$, but I don't see how that helps. Jan 16, 2020 at 0:00
• It would help to correctly draw the picture. On your picture, $E$ is on the segment $AC$ but not on the segment $BD$.
– user700480
Jan 16, 2020 at 0:12
• Oh. Let me try that. Jan 16, 2020 at 0:12
• Ok. It was 135. Thanks @StinkingBishop! Jan 16, 2020 at 0:16
• Glad to help. Feel free to answer your own question then - someone may find it handy in the future, when they browse this Web site.
– user700480
Jan 16, 2020 at 0:17

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 $$\angle C+\angle D=135^\circ$$.