, 2019

Consider any two points $F, G$ on $\omega$ such that $\angle BFD = \angle DGC$. Exploiting the isosceles triangles $\triangle AFG$, $\triangle AFD$, and $\triangle ADG$, we deduce (using directed angles throughout):
$$\angle DBF - \angle GCD = 180^\circ - \angle BFD - \angle BDF - (180^\circ - \angle DGC - \angle CDG) \stackrel{*}{=} \\ \angle CDG - \angle FDB = \frac{1}{2} \cdot (\angle DAG - \angle DAF) = \frac{1}{2} \cdot [(180^\circ - 2 \angle ADG) - (180^\circ - 2 \angle ADF)] = \\ \angle ADF - \angle GDA = \angle DFA - \angle AGD = \angle DFG - \angle FGD \stackrel{*}{=} \angle BFG - \angle FGC,$$
where we use $\angle BFD = \angle DGC$ at $(*)$. Thus $BFGC$ is cyclic.

Figure 3: G3

Now, if in addition $FG \perp AO$, then since $A$ is the centre of $\omega$, in fact $AO$ is the perpendicular bisector of $FG$. But by definition, since $ABC$ is scalene, $AO$ meets the perpendicular bisector of $BC$ at $O$. Hence $O$ is the centre of $BFGC$, and thus in fact $BFAGC$ is cyclic. But then the lines perpendicular to $AF$ at $F$, and $AG$ at $G$ (the tangents to $\omega$) must intersect at $E$, the point antipodal to $A$ on $\odot BFAGC$. $\square$

Let the circumcircle of $ABC$ be $\Gamma$. From the conditions, $G$ is the reflection of $F$ in the line $AO$. Let $B', D'$ be the reflections of $B, D$ across this same line $AO$. Clearly $D'$ also lies on $\omega$ and $B'$ lies on $\Gamma$.
Then, using directed angles, $\angle CGD = \angle DFB = \angle B'GD'$ so
$$\angle B'GC = \angle B'GD' - \angle CGD' = \angle CGD - \angle CGD' = \angle D'GD = \frac{1}{2}\angle D'AD = \angle OAD.$$

Then, exploiting the isogonality property that $\angle DAB = \angle CAO$, we have
$$\angle OAD = \angle CAB - 2\angle DAB = \angle ABC - \angle BCA = \angle ABC - \angle B'BA = \angle B'BC.$$
So $G$ lies on $\Gamma$, and by the reflection property so does $F$.
But then, as in the previous solution, the tangents at $F$ and $G$ to $\omega$ must intersect at $E$, the point antipodal to $A$ on $\Gamma$. $\square$