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Question: Find $\angle BDE$ when $\angle A=20^\circ$, $\overline{AB}=\overline{AC}$, $\angle CBD=65^\circ$, and $\angle BCE=25^\circ$.

Figure So far I figured out that $\angle BDE=\angle AED-15^\circ$ and $\overline{BD}\bot\overline{CE}$ that don't seem to be helping.

Kay K.
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    See http://math.stackexchange.com/questions/750410/worlds-hardest-easy-geometry-problem – Robert Z Nov 25 '16 at 07:04
  • @RobertZ Thanks a lot..! – Kay K. Nov 25 '16 at 07:22
  • @RobertZ I thought the answer that got the most votes to the post you mentioned was the answer you wanted me to look at but when I tried it I found it doesn't solve the problem (someone commented it gives only a linearly dependent equation system that's same to what I found). I don't see that the post you gave me has any real answer. – Kay K. Nov 25 '16 at 07:40
  • Try the external link: http://thinkzone.wlonk.com/MathFun/Triangle.htm This is a famous problem. – Robert Z Nov 25 '16 at 07:42
  • @RobertZ Thanks. I still don't see any solution there. The website owner is offering that if someone sends an e-mail then he might send back the answer. However, it seems that the solution is dependent to specific angles, but those two problems in the website have different angles than this one. Did you actually take a look at the external page and see what it's saying? Do you know the solution?? – Kay K. Nov 25 '16 at 07:57
  • I gave a solution using trigonometry. I hope it helps. – Robert Z Nov 25 '16 at 09:18
  • Search for "adventitious angles". – Gerry Myerson Nov 25 '16 at 09:33

2 Answers2

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This is a variation of the notorious Langley's problem.

We follow a trigonometric approach. Let $\angle BDE=x$.

i) Use the Law of sines in $\triangle EBC$, $$\frac{\sin 25^\circ}{EB}=\frac{\sin75^\circ}{AB}.$$

ii) Use the Law of sines in $\triangle DBC$, $$\frac{\sin 80^\circ}{BD}=\frac{\sin 35^\circ}{AB}.$$

iii) Use the Law of sines in $\triangle EBD$, $$\frac{\sin x}{EB}=\frac{\sin (165^\circ-x)}{BD}=\frac{\sin (15^\circ+x)}{BD}.$$

It follows that $$\frac{\sin (15^\circ+x)}{\sin x}=\frac{\sin 80^\circ}{\sin 35^\circ}\cdot \frac{\sin 75^\circ}{\sin 25^\circ}$$ Since $\sin (15^\circ+x)=\sin 15^\circ\cos x+\cos 15^\circ\sin x$, we obtain that $$\cot x=\frac{1}{\sin 15^\circ}\cdot\frac{\sin 80^\circ}{\sin 35^\circ}\cdot \frac{\sin 75^\circ}{\sin 25^\circ}-\cot 15^\circ.$$ Therefore $x=5^\circ$.

Robert Z
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This is the general solution for any angle: enter image description here

Seyed
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