KöMaL 2009 A495

In the acute triangle ABC we have $\angle$ BAC =$\alpha$. The point $D$ lies in the interior of the triangle, on the bisector of $\angle$ BAC , and points $E$ and $F$ lie on the sides $AB$and $BC$, respectively, such that $\angle$ BDC =2$\alpha$, $\angle$ AED $=90^o$+$\frac{\alpha}{2}$, and $\angle$ BEF =$\angle$ EBD . Determine the ratio $BF$:$FC$.