Let $ABC$ be isosceles triangle $(AB=AC)$ with $\angle BAC = 20^{\circ}$. Point $D$ is on side $AC$ such that $\angle DBC = 60^{\circ}$. Point $E$ is on side $AB$ such that $\angle ECB = 50^{\circ}$. Find $\angle EDB $
Solution:
Let $x=\angle EDB $. We see that $BC = BE$
By sinus theorem for $BCE$ and $BCD$ we have $${\sin x\over \sin (x+20^{\circ})}={BE\over BD} = {\sin 40^{\circ}\over \sin 80^{\circ}} = {1\over 2\cos 40^{\circ}} $$ From here we get $$\tan x = {\sin 20^{\circ}\over 2\cos 40^{\circ}-\cos 20^{\circ}}$$ and it is not difficult to see (after some algebraic manipulation) that $x = 30^{\circ}$
Here is a question: Is there a neat synthetic soultion to this problem?
