Mediterranean Mathematical Olympiad

Let $O$ be the center of the circle $\Gamma$ and $R$ its radius. Let $G$ be the other point of intersection of the circles $FDC$ and $BCE$. (See Figure 1).

Remark. The following argument needs some minor changes if all the angles of $ABCD$ are acute, that is, if the center $O$ of $\Gamma$ is internal to $ABCD$.

$\angle FGC = \angle FDC$ because both are inscribed angles in $FDC$ subtending the same arc. $\angle EGC = [EGCB \text{ is cyclic}] = 180^\circ - \angle CBE = [ABCD \text{ is cyclic}] = 180^\circ - \angle CDF = 180^\circ - \angle CGF$. Then we have $\angle EGC + \angle CGF = 180^\circ$, and so $EG$ and $GF$ lie on the same straight line.

For the second part of the argument, we consider Figure 2:

Let $L$ be the midpoint of $EF$ and draw the tangents (in red in the figure) to the circle $\Gamma$ from $F$ (tangency point $H$), $E$ (tangency point $K$) and $L$ (tangency point $M$). The triangles $FGC$ and $FEB$ are similar (they have the same angles); then we have
$$\frac{FG}{FC} = \frac{FB}{FE} \Leftrightarrow FG \cdot FE = FC \cdot FB = FH^2$$
on account of the power of $F$ with respect to $\Gamma$. Likewise, we have
$$EG \cdot EF = EC \cdot ED = EK^2$$
Adding up the preceding, yields
$$FG \cdot FE + EG \cdot EF = EF(FG + GE) = EF^2 = FH^2 + EK^2$$
On account that $EF = 2 \cdot FL$, the last equality can be written as
$$4 \cdot FL^2 = FH^2 + EK^2,$$
which can be transformed into
$$4 \cdot FL^2 + 2R^2 = FH^2 + R^2 + EK^2 + R^2.$$
Since $FH^2 + R^2 = FO^2$, and $EK^2 + R^2 = EO^2$, then we get
$$4 \cdot FL^2 + 2R^2 = FO^2 + EO^2$$
But $FO^2 = FL^2 + LO^2$, and $EO^2 = LE^2 + LO^2 = FL^2 + LO^2$ because $FL = FE$. Then we obtain
$$4 \cdot FL^2 + 2R^2 = 2OL^2 + 2FL^2 \Rightarrow FL^2 + R^2 = OL^2 = LM^2 + R^2,$$
and this means $FL = FM$. But then the circle of center $L$ and radius $LE = LF$ passes through $M$, and the radii $LM$ and $MO$ are perpendicular to each other, and we are done. $\square$