Problem 1827

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
$$\begin{equation*} \angle KFA = \angle AGL . \tag{1} \end{equation*}$$
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.

Again, we denote the circumcircle of $BDKF$ by $\omega_B$. In addition, we set $\alpha= \angle BAC$, $\varphi=\angle ABF$, and $\psi=\angle EDA=\angle AED$ (see Figure 2). Notice that $AF=AG$ entails $\varphi=\angle GCA$, so all three of $\alpha, \varphi$, 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
$$2 \angle KFA = \angle BFA + \varphi - \psi - \angle CBA .$$
Now the quadrilateral $AFBC$ is cyclic, so this entails $2 \angle KFA = \alpha + \varphi - \psi$.
Due to the "symmetry" between $B$ and $C$ alluded to above, this argument also shows that $2 \angle AGL = \alpha + \varphi - \psi$. This concludes the proof of (1).

Figure 2