BMO 2019 Shortlist

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

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Alternative solution.

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

Then, exploiting the isogonality property that $\angle DAB=\angle CAO$, we have $$\angle OAD=\angle CAB-2\angle DAB=\angle ABC-\angle BCA=\angle ABC-\angle B^{\prime }BA=\angle B^{\prime }BC.$$ 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 $\Gamma$. $\square$