Saudi Arabian Mathematical Competitions

Denote $O$ as the projection of $B$ on $AC$ and $BK$ is the diameter of $(O, OB)$. It is easy to see that $E \in (O)$. We have
$$\angle BCA = \angle AEB - \angle CBE = \angle OBE - \angle ABE = \angle OBA = \angle OKA.$$
This implies that $ODCK$ is the inscribed quadrilateral and $\angle KDC = 90^{\circ}$. Hence, $K, A, D$ are collinear and $D \in (O)$.
Triangle $CBK$ is isosceles then $CE$ is the angle bisector of $\angle BCK$. Note that $DE$ is the angle bisector of $\angle ADC$. Then $E$ is the incenter of triangle $KDC$.

So $O_2$ is the midpoint of minor $\operatorname{arc} CD$ of the circumcircle of $KDC$, then $K, E, O_2$ are collinear.
We have
$$\triangle BAD \sim \triangle KCD \Rightarrow \frac{BD}{BA} = \frac{KD}{KC}.$$
Based on the property of bisector in triangle, we have
$$\frac{FD}{FA} = \frac{BD}{BA} = \frac{KD}{KC} = \frac{HD}{HC}.$$
Hence, $FH \parallel AC$. We have $O_1$ is the midpoint of $FH$ also is the circumcenter of quadrilateral $DFEH$.
From this, we can conclude that $EG$ is the angle bisector in triangle $BEH$ and
$$\frac{GH}{GB} = \frac{EH}{EB} = \tan \frac{B}{2} = \tan \frac{\alpha}{2}.$$