Russian Mathematical Olympiad

Let $M$ be the midpoint of side $AC$, the line $BM$ intersects the circle $\Omega$ again at point $F'$, and the line $FM$ intersects the circle $\Omega$ again at point $B'$ (see Fig. 20).

Since arcs $AE$ and $CE$ are equal, points $M$ and $E$ lie on the perpendicular bisector $\ell$ of segment $AC$. Therefore, $\angle EMD = 90^\circ$ and, consequently, $M$ lies on the circle $\omega$.

We have:
$$\frac{1}{2}\overline{B'C'E} = \angle B'FE = \angle MFE = \angle MDE = \angle CDE = \frac{1}{2}(\overline{AB} + \overline{CE}) = \frac{1}{2}(\overline{AB} + \overline{AE}) = \frac{1}{2}\overline{BAE}.$$
From the equality of arcs $B'C'E$ and $BAE$, it follows that points $B$ and $B'$ are symmetric with respect to $\ell$, so lines $BM$ and $B'M$ (and therefore points $F'$ and $F$) are symmetric with respect to $\ell$. The last statement means that $\overline{EF} = \overline{EF'}$, whence $\angle FBE = \angle F'BE$. We obtain that lines $BF$ and $BM$ are symmetric with respect to line $BE$, which is what was required to be proved.

Let $M$ be the midpoint of side $AC$. As in the previous solution, we obtain that $ME$ is the perpendicular bisector of $AC$, and $M$ lies on $\omega$. Let the line $DF$ intersect the circle $\Omega$ again at point $G$. Since $\angle DFE = 90^\circ$, $G$ is the point diametrically opposite to $E$; in particular, $EG$ passes through $M$ (see Fig. 21). We have $\angle FBE = \angle FGE$.

Furthermore, since $EG$ is a diameter, $\angle GBE = 90^\circ$. From the equalities $\angle GBD = \angle GMD = 90^\circ$, it follows that $GBDM$ is a cyclic quadrilateral (with diameter $DG$), whence $\angle MBE = \angle MBD = \angle MGD = \angle EGF$. Thus, $\angle FBE = \angle FGE = \angle MBE$, which is what was required to establish.