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 $\mathrm{arc}CD$ of the circumcircle of $KDC$, then $K,E,O_2$ are collinear. We have $$△BAD\sim △KCD\Rightarrow \frac{BD}{BA}=\frac{KD}{KC}.$$ 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 $$\frac{GH}{GB}=\frac{EH}{EB}=\tan \frac{B}{2}=\tan \frac{\alpha }{2}.$$