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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

Drawing of question

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): enter image description here

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  • $\begingroup$ I can see that it's $\angle ABE+90^{\circ}$, but I don't see how that helps. $\endgroup$
    – asdf334
    Jan 16, 2020 at 0:00
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    $\begingroup$ It would help to correctly draw the picture. On your picture, $E$ is on the segment $AC$ but not on the segment $BD$. $\endgroup$
    – user700480
    Jan 16, 2020 at 0:12
  • $\begingroup$ Oh. Let me try that. $\endgroup$
    – asdf334
    Jan 16, 2020 at 0:12
  • $\begingroup$ Ok. It was 135. Thanks @StinkingBishop! $\endgroup$
    – asdf334
    Jan 16, 2020 at 0:16
  • $\begingroup$ 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. $\endgroup$
    – user700480
    Jan 16, 2020 at 0:17

1 Answer 1

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

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