56th International Mathematical Olympiad Shortlisted Problems

It suffices to prove that the lines $FK$ and $GL$ are symmetric about $AO$. Now the segments $AF$ and $AG$, being chords of $\Omega$ with the same length, are clearly symmetric with respect to $AO$. Hence it is enough to show $$\begin{array}{c}\angle KFA=\angle AGL.\end{array}\text{\hspace{1em}(1)}$$ Let us denote the circumcircles of $BDF$ and $CEG$ by ${\omega }_B$ and ${\omega }_C$, respectively. To prove (1), we start from $$\angle KFA=\angle DFG+\angle GFA-\angle DFK.$$ In view of the circles ${\omega }_B$, $\Gamma$, and $\Omega$, this may be rewritten as $$\angle KFA=\angle CEG+\angle GBA-\angle DBK=\angle CEG-\angle CBG.$$ Due to the circles ${\omega }_C$ and $\Omega$, we obtain $\angle KFA=\angle CLG-\angle CAG=\angle AGL$. Thereby the problem is solved.

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Alternative solution.

Again, we denote the circumcircle of $BDKF$ by ${\omega }_B$. In addition, we set $\alpha =\angle BAC$, $\phi =\angle ABF$, and $\psi =\angle EDA=\angle AED$ (see Figure 2). Notice that $AF=AG$ entails $\phi =\angle GCA$, so all three of $\alpha ,\phi$, and $\psi$ respect the "symmetry" between $B$ and $C$ of our configuration. Again, we reduce our task to proving (1). This time, we start from $$2\angle KFA=2(\angle DFA-\angle DFK).$$ Since the triangle $AFD$ is isosceles, we have $$\angle DFA=\angle ADF=\angle EDF-\psi =\angle BFD+\angle EBF-\psi .$$ Moreover, because of the circle ${\omega }_B$ we have $\angle DFK=\angle CBA$. Altogether, this yields $$2\angle KFA=\angle DFA+(\angle BFD+\angle EBF-\psi )-2\angle CBA,$$ which simplifies to $$2\angle KFA=\angle BFA+\phi -\psi -\angle CBA.$$ Now the quadrilateral $AFBC$ is cyclic, so this entails $2\angle KFA=\alpha +\phi -\psi$. Due to the "symmetry" between $B$ and $C$ alluded to above, this argument also shows that $2\angle AGL=\alpha +\phi -\psi$. This concludes the proof of (1).

Figure 2