SAUDI ARABIAN MATHEMATICAL COMPETITIONS

1) First, notice that $\angle AFB=\angle AFE+\angle BFE=\angle ADB+\angle ACB=\angle AOB$, then $A,F,O,B$ are concyclic. Similarly, $D,F,O,C$ are also concyclic.

By considering three radical axes of three circles $(O)$, $(ABOF)$, $(CDFO)$, we can see that three lines $AB$, $CD$, $FO$ are concurrent at $S$ which implies that $S,F,O$ are collinear. We have $$\begin{array}{rl}\angle EFO& =\angle AFO-\angle AFE=180^{\circ }-\angle ABO-\angle ADB\\ & =180^{\circ }-\angle ABO-\frac{1}{2}\angle AOB=90^{\circ }.\end{array}$$ Then $SF\perp EF$.



2) Suppose that the angle bisectors of $\angle GAF$, $\angle GBF$ intersect at $I$. We shall prove that $I,E,F$ are collinear. Indeed, we have $$\begin{array}{rl}& \angle IAG=\angle IAF\iff \angle GAB+\angle IAB=\angle IAE+\angle EAF\\ & \iff \angle FDC+\angle EDF+\angle IAB=\angle IAE+2\angle EAF\\ & \iff \angle EDC-\angle IAE+\angle IAB=2\angle EAF\\ & \iff 2\angle IAB=2\angle EAF\iff \angle IAB=\angle EAF\end{array}$$ Hence, $IA$, $EA$ are isogonal conjugate in angle $\angle BAF$. Similarly, $IB$, $EB$ are isogonal conjugate in angle $\angle ABF$. Then $IF$, $EF$ also are isogonal conjugate in angle $\angle AFB$. But $EF$ is the bisector of angle $\angle AFB$ then $I,E,F$ are collinear and $IF$ is the angle bisector of $\angle AFB$.

In quadrilateral $GAFB$, we have three internal angle bisectors are concurrent at $I$ then this quadrilateral is circumscribed, which means $GA+FB=GB+FA$. $\square$