Call $k$ the $G$ median of triangle $GHK.$ Let $R$ and $S$ be the midpoints of the smaller arcs $AB$ and $AC$. Because $(O)$ and $(EMB)$ are orthogonal, we have that $SP$ is a diameter of $(O),$ hence $P$ is the midpoint of bigger arc $AC.$ Analogously, $Q$ is the midpoint of the bigger arc $AB.$ Let $l$ be a line passing through $G$ parallel to $EF.$ Let $I_A$ and $J$ be the $A$ excenter of triangle $ABC$ and the midpoint of smaller arc $BC$ respectively. It is know that $OI_A \perp EF,$ thus $OI_A \perp l.$ Let $m$ be the parallel from $O$ to the $A$ bisector of $ABC.$ Clearly $(OI_A, OI, OJ, m)=-1$. Because $OI_A \perp l,$ $OJ \perp BC,$ $m\perp PQ,$ and $G(l,k,BC,PQ)=-1$ it follows that $OI$ is also perpendicular to $k.$

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Let $I_a$, $I_b$, $I_c$ be $A$ - excenter, $B$ - excenter, $C$ - excenter of $\triangle$ $ABC$; $J$ be center of ($I_aI_bI_c$); $H'$ be intersection of tangents at $A$ of ($O$) and $PQ$; $G'$ $\equiv$ $I_aG$ $\cap$ $I_bI_c$; $T$ be midpoint of $GG'$ It's easy to see that: $BM$, $CN$ are $B$ - symmedian, $C$ - symmedian of $\triangle$ $ABC$ So: $\dfrac{MC}{MA}$ = $\dfrac{BC}{AB}$ = $\dfrac{EC}{EA}$ or $ME$ is internal bisector of $\widehat{CMA}$ Then: $P$ is midpoint of $\stackrel\frown{ABC}$ Similarly: $Q$ is midpoint of $\stackrel\frown{ACB}$ Since: $P_{H' / (I_aI_bI_c)}$ = $H'I_a^2$ = $H'A^2$ = $\overline{H'P}$ . $\overline{H'Q}$ = $P_{H' / (O)}$, we have: $H'$ lies on radical axis of ($I_aI_bI_c$) and ($O$) or $KH'$ $\perp$ $IO$ We also have: $P_{E / (II_bI_c)}$ = $\overline{EI}$ . $\overline{EI_b}$ = $\overline{EC}$ . $\overline{EA}$ = $P_{E / (O)}$ and $P_{F / (II_bI_c)}$ = $\overline{FI}$ . $\overline{FI_c}$ = $\overline{FA}$ . $\overline{FB}$ = $P_{F / (O)}$ So: $EF$ is radical axis of ($O$) and ($II_bI_c$) or $I_aO$ $\perp$ $EF$ But: $P_{G / (BICI_a)}$ = $\overline{GB}$ . $\overline{GC}$ = $\overline{GP}$ . $\overline{GQ}$ = $P_{G / (PJQI_a)}$ then $I_aG$ is radical axis of ($BICI_a$) and ($PJQI_a$) or $I_aG$ $\perp$ $IO$ We have: ($I_aJ$; $I_aI$) $\equiv$ ($I_aJ$; $I_aI_c$) + ($I_aI_c$; $I_aI$) $\equiv$ $\dfrac{\pi}{2}$ $-$ ($I_bI_c$; $I_bI_a$) + $\dfrac{\pi}{2}$ $-$ ($I_cI_b$; $I_cI_a$) $\equiv$ ($I_cI_a$; $I_cI_b$) $-$ ($I_bI_c$; $I_bI_a$) $\equiv$ ($KG$; $KG'$) (mod $\pi$) and ($JI$; $JI_a$) $\equiv$ ($GG'$; $GK$) (mod $\pi$) Hence: $\triangle$ $JI_aI$ $\stackrel{+}{\sim}$ $\triangle$ $GKG'$ But: $O$, $T$ are midpoint of $JI$, $GG'$, respectively so: $\triangle$ $I_aOI$ $\stackrel{+}{\sim}$ $\triangle$ $KTG'$ Then: ($OI$; $OI_a$) $\equiv$ ($TG'$; $TK$) (mod $\pi$) Combine with: $I_aT$ $\perp$ $IO$, we have: $I_aO$ $\perp$ $KT$ or $K$, $H$, $E$, $F$, $T$ are collinear We also have: $PQ$ $\parallel$ $I_bI_c$ Hence: $T$ is midpoint of $KH$ Therefore: $GT$ $\perp$ $IO$ of $G$ - median of $\triangle$ $KGH$ perpendiculars to $IO$