A problem from BdMO 2013:
Let $ABC$ be an isoscles triangle with $AB=AC$.The bisector of $\angle B$ meets $AC$ at $D$.Given that $BC=BD+AD$,we need to figure out $\angle A$.
If we consider $\angle B=\angle C=\angle 2x$,then after bisecting angle B,we get a triangle with angles equal to $x$,$2x$,and $180-3x$.But that does not get us any further except that $\angle ADB=3x$.I also tried extending BD to $A'$ such that $A'D=AD$ but that does not help at all.Finally,I tried to utilize the Angle Bisector Theorem but that yielded nothing good as well.A prod in the correct direction would be appreciated.
NOTE: I am looking for a hint,not the whole solution.