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
$$\begin{aligned} \angle EFO & = \angle AFO - \angle AFE = 180^\circ - \angle ABO - \angle ADB \\ & = 180^\circ - \angle ABO - \frac{1}{2} \angle AOB = 90^\circ . \end{aligned}$$
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
$$\begin{aligned} & \angle IAG = \angle IAF \Leftrightarrow \angle GAB + \angle IAB = \angle IAE + \angle EAF \\ & \Leftrightarrow \angle FDC + \angle EDF + \angle IAB = \angle IAE + 2 \angle EAF \\ & \Leftrightarrow \angle EDC - \angle IAE + \angle IAB = 2 \angle EAF \\ & \Leftrightarrow 2 \angle IAB = 2 \angle EAF \Leftrightarrow \angle IAB = \angle EAF \end{aligned}$$
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 $GA + FB = GB + FA$. $\square$