Czech-Slovak-Polish Match

Clearly $DF$ is a diameter of $k$. First we show that under the given conditions the vertex $C$ cannot lie in the half-plane $DFA$. If the vertices $B$, $C$ are points on the subarc $DA$ of the arc $DAF$ (Fig. 1) then the angles $DCB$ and $DBA$ are obtuse, hence $|DC|<|DB|<|DA|<|DE|$, which contradicts to the equality $|CD{|}^2=|AD|\cdot |ED|$. If the vertices $B$, $C$ are points on the subarc $AF$ of the arc $DAF$ (Fig. 2) the angle $BAE$ is acute and $|\angle DBE|=180^{\circ }-|\angle DBC|\le 90^{\circ }$, so the possible other meeting point $B^{\prime }$ of the half-line $DB$ with the circumcircle of the triangle $AEB$ lies in the segment $DB$. Hence $|DC|>|DB|\ge |D{B}^{\prime }|$. This means that the equality $|CD{|}^2=|AD|\cdot |ED|$ cannot hold as $|AD|\cdot |ED|=|DB|\cdot |D{B}^{\prime }|$ (which is the power of $D$ with respect to the circumcircle of the triangle $AEB$).

Fig. 1

Fig. 2

We have shown that the vertex $C$ of the given quadrangle does not lie in the half-plane $FDA$, hence $|FC|=|DA|$ if and only if $DAFC$ is a rectangle, i.e. if and only if $CA$ is a diameter of the circle $k$, which is equivalent to the angle $CBA$ being right, which is in turn equivalent to the triangle $AEB$ being right with the right angle at $B$, i.e. to the circumcenter of the triangle $AEB$ being the midpoint of $AE$.