We know $AD$ is the $A$-symmedian of $\triangle ABC$, so it is well known that $B,C,D,O$ are concyclic. So $(BCD)=(BCO)$. Then $\angle EAO = \angle CAO=90-B$, and $\angle AEF=180-\angle CEF=180-\angle CBF=B$ since $E,C,F,B$ are concyclic. Therefore, $AO\perp EF$.

It's not important that $D$ is defined as the intersection of the tangents. We can replace the circumcircle of $\triangle BCD$ with any circle that passes through $B$ and $C$. Indeed, we know that $\angle CAO = 90^\circ - \angle B$, and since $BCEF$ is cyclic, we have that $\angle B = \angle CEF$, from which the result follows.

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Observe that we can obtain $\Delta AEF$ by reflecting $\Delta ABC$ about its $A-$angle bisector and taking a homothety at $A$. Now since the circumcentre and orthocentre are isogonal conjugates, $O$ has to lie on the $A-$altitude of $\Delta AEF$, and therefore we are done. $\square$

Let $H$ be the orthocenter of $ABC$. $\measuredangle BEC = \measuredangle BDC = \measuredangle BFC \implies BCEF$ is concyclic $\implies \Delta ABC \sim \Delta AEF$. Consider taking a suitable homothety about A then reflecting across the angle bisector of $\angle BAC$, such that $BC$ goes to $EF$. Under this transformation, line $AH$ goes to line $AO$ because they are isogonal in $\angle BAC$. Since $AH \perp BC$, we have $AO \perp EF$.

We know that $\measuredangle OAC=90^{\circ}-\measuredangle CBA=\measuredangle KAE$ Also, $\measuredangle CBA=\measuredangle CEF=\measuredangle AEK$ Now, $\measuredangle KAE+\measuredangle AEK+\measuredangle EKA=0$ $\implies 90^{\circ}-\measuredangle CBA+\measuredangle CBA +\measuredangle EKA=0$ $\implies \measuredangle EKA=90^{\circ}$ And we are done!

Claim: $O$ lies on the circumcircle of $\triangle BDC$ proof: $$\measuredangle BAC =\measuredangle BCD \implies \measuredangle BDC= 180^{\circ}- 2\times \measuredangle BAC$$, we know that $$\measuredangle BOC = 2\times \measuredangle BAC$$so $$OBDC \text{ is cyclic quad }$$ $$\measuredangle FBC=\measuredangle AEF $$$$\measuredangle OAC=\measuredangle OCA$$$$\measuredangle OCB=90^{\circ}$$we know that $$\measuredangle AEG=90^{\circ}-\measuredangle BAO-\measuredangle OAE + \measuredangle BAO\implies 90^{\circ}-\measuredangle OAE$$$$\measuredangle AGE =180^{\circ}-(90^{\circ}-\measuredangle OAE+\measuredangle OAE)=90^{\circ}$$hence we are done $\blacksquare$

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