Saudi Arabia booklet, 2024

1) Let $T$ be the intersection of the tangent at $A$ of $(O)$ with $BC$. Then, because $ABDC$ is a harmonic quadrilateral, $TD$ is also tangent to $(O)$. It follows that $A, D, G$ both belong to the circle of center $T$, which is also the Apollonius circle at vertex $A$ of triangle $ABC$. On the other hand, $(T)$ and $(O)$ are orthogonal so $OA, OD$ are tangent to $(T)$, then it is easy to check that $GO$ is the symmedian of triangle $ADG$. Let $K$ be the second intersection of $AG$ and $(O)$. Then,
$$\angle BEK = \angle BAK = \angle BGK \implies KE = KG.$$
Similarly, $KF = KG$ so $KE = KF$ implies that $K$ is the midpoint of the minor arc $EF$ of $(O)$. It follows that $AG$ is the angle bisector of $\angle EAF$.

Let $M$ be the midpoint of $BC$ and take $A' \in (O)$ such that $AA' \parallel BC$ then since $MA' = MA = MG$ implies that $G, M, A'$ are collinear. On the other hand, since
$$A'(AD, BC) = -1 \text{ and } AA' \parallel BC,$$
we have $A', M, D$ are collinear. From there, it can be deduced that $GD$ passes through $A'$. Since $K, O, A'$ are also collinear, $A'$ is the midpoint of the major arc $EF$ of the circle $(O)$. Then, $K$ is the center of the circumcircle of $GEF$ and $A'E \perp KE$, $A'F \perp KF$ so $A'$ is the intersection point of two tangent lines of $(K)$ at $E, F$. Therefore, $GD$ is the symmedian of triangle $GEF$.

2) We will prove that $D$ is the tangent point of the external Mixlinear circle with respect to vertex $A$ in triangle $AEF$. Indeed, let $D', U, V$ be the tangent points of the external Mixtilinear circle $(\omega)$ of triangle $AEF$ with $(O)$, $AE, AF$ respectively. Then, $G$ is the midpoint of $UV$, and $D'U, D'V$ respectively pass through the midpoints of the arc $AE$ (contains $F$) and $AF$ (contains $F$) of $(O)$. Note that
$$\angle ACB = \angle GCB = \angle BCF = \angle BAF$$
so $B$ is the midpoint of the arc $AE$ (contains $F$) of $(O)$ so $B, D', V$ are collinear. Similarly, $C, D', U$ are collinear. Hence, we have $UV \parallel BC$. Since $(\omega)$ and $(O)$ are tangent, so we have
$$\angle FD'V = \angle FCD' + \angle D'UV = \angle FCD' + \angle D'CB = \angle BCF = \angle FGV$$
implies that $D'FVG$ is cyclic. Similarly $D'EUG$ is also cyclic. Let $D'$ be the opposite ray of $D'G$ then
$$\angle FD'x = \angle FVG = \angle AVU = \angle AUV = \angle EDx$$
so $D'$ passes through the midpoint of the major arc $EF$ of $(O)$, which is $A'$. From here, $D \equiv D'$. Finally, let $I$ be the incenter of triangle $AEF$ and consider $\Omega$ to be the inversion of center $A$ and power $AE \cdot AF$, union to the reflection about $AG$. Then, according to the properties of this transformation which preserve the tangency and the isogonal properties, one can get:
$$\begin{gathered} \Omega : (I) \to (\omega) \text{ and } \Omega : BC \leftrightarrow (O) \\ \implies \Omega : J \leftrightarrow D. \end{gathered}$$