Question 162
Let $ABC$ be a triangle in which $A$ is an obtuse angle, and let $\alpha = \angle ABC$. Denote by $BE$ the internal angle bisector of triangle $ABC$ with $E \in AC$, and suppose that $\angle AEB = 45^{\circ}$. The altitude $AD$ of triangle $ABC$ intersects $BE$ at $F$. Let $O_1, O_2$ be the circumcenters of triangles $FED$, $EDC$. Suppose that $EO_1, EO_2$ meet $BC$ at $G, H$ respectively. Prove that $\frac{GH}{GB} = \tan \frac{\alpha}{2}$.