Saudi Booklet

Suppose $EF$ intersects $BC$ at $D'$. Let $(\omega_A)$ be the circle with diameter $DD'$. By the angle bisector theorem, one can get
$$\frac{DB}{DC} = \frac{AC}{AB}, \quad \frac{EC}{EA} = \frac{AB}{BC}, \quad \frac{FA}{FB} = \frac{BC}{CA}.$$
Based on Ceva's theorem, $AD, BE, CF$ are concurrent and $(DD', BC) = -1$. If $AU, AV$ are the internal, external angle bisectors of triangle $ABC$, then $(UV, BC) = -1$. Through the reflection over the center $M$, then $(DV', CB) = -1$ with $V'$ being the symmetric point to $V$ through $M$. From there, $D' \equiv V'$ so $(\omega_A)$ is the reflection of the $A$-Apollonius circle through the perpendicular bisector of $BC$.

Suppose that $A$-Apollonius circle intersects $(O)$ at $K$ then $\frac{KB}{KC} = \frac{AB}{AC}$ so $ABKC$ is harmonic, entailing $AK$ is the symmetric line of triangle $ABC$. Therefore, $(\omega_A)$ intersects $(O)$ at $A', R$ will be the points symmetric to $A, K$ through the perpendicular bisector $BC$. Therefore, $KR \parallel BC$ so $AK, AR$ are isogonal in angle $A$, which implies that $AR$ is the median of triangle $ABC$ or $R \in AM$. Suppose $AR$ intersects $(\omega_A)$ at $S$, then by the property of power of a point and Newton's formula, we have
$$\overline{MR} \cdot \overline{MS} = \overline{MD} \cdot \overline{MD'} = MB^2 = -\overline{MB} \cdot \overline{MC} = -\overline{MR} \cdot \overline{MA}.$$
From there, we deduce $\overline{MS} = -\overline{MA}$ or $M$ is the midpoint of $AS$. Here we have
$$\frac{\mathcal{P}_{G/(O)}}{\mathcal{P}_{G/(\omega_A)}} = \frac{\overline{GR} \cdot \overline{GA}}{\overline{GR} \cdot \overline{GS}} = \frac{\overline{GA}}{\overline{GS}} = \frac{2/3 \cdot \overline{MA}}{4/3 \cdot \overline{MS}} = -\frac{1}{2}.$$
Therefore $\mathcal{P}_{G/(\omega_A)} = -2\mathcal{P}_{G/(O)}$ is equal between vertices $A, B, C$. Therefore we have $G$ has the same power to the three circles $(\omega_A), (\omega_B), (\omega_C)$.

On the other hand, $OA$ is tangent to the Apollonius circle at vertex $A$ so $OA'$ is tangent to $(\omega_A)$. From there we have
$$\mathcal{P}_{O/(\omega_A)} = OA'^2 = R^2$$
with $R$ being the radius of $(O)$. Also because of this symmetry, $G$ has the same power to the three circles $(\omega_A), (\omega_B), (\omega_C)$. Let $X$ be the projection of $D$ onto $EF$ then $X \in DH$ and $X \in (\omega_A)$. Then, $\mathcal{P}_{H/(\omega_A)} = \overline{HX} \cdot \overline{HD}$. Obviously, this quantity is also symmetric between the vertices $D, E, F$ in triangle $DEF$, so $H$ also has the same power to the three circles $(\omega_A), (\omega_B), (\omega_C)$. From that we immediately have $H \in OG$.

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