Triangle $ABC$ is isosceles ($CA=CB$). $BD$ is angle bisector of $\angle B$. Find angles of triangle $ABC$ if $BD+DC=AB$.
Actually, I have a solution but I don't like it too much:
If you apply law of sines to $\triangle BCD$, you get:
Same thing for $\triangle ABD$:
Eliminating $x$ is easy but you'll have to simplify the resulting equation. After about an hour I got something that looked like a good starting point:
But I almost gave up here because if you try to expand multiple angle items like $\cos2\alpha$ or $\sin3\alpha$, you get a nasty equation that cannot be solved. However, my last attempt succeeded... I replaced $(\sin3\alpha-\sin2\alpha)$ with product, cancelled common factors on both sides, then replaced product on the left side with sum and eventually got the following:
So much work for such a "nice" angle! Considering the simplicity of the result I doubt that there has to be a simpler way to solve this problem without trignonometry.
However, I was not able to find it and so I decided to share this problem with you.