BMO Short List

Let $K$ be the intersection point of $ZE$ and $AC$. Then $K$ lies on the radical axis of the two circles, so it has equal powers to both circles. The power of the point $K$ with respect to the one circle is $KA\cdot KC$, while the power to the other circle is $R^2-K{C}^2=C{D}^2-K{C}^2$.

From the theorem of Pythagoras in triangle $ADC$ we get $A{C}^2=2C{D}^2$. We set $KA=x$, $KC=y$, and combining all the above yields $xy=\frac{(x+y{)}^2}{2}-y^2$, hence $2xy=x^2+y^2+2xy-y^2$, i.e. $x=y$.

This means that $K$ is the midpoint of $AC$. Moreover, we have $\angle ADC=\angle AMC=90^{\circ }$, so the points $A$, $D$, $M$, $C$ are on the same circle – let it be $(c_1)$ – and the center of this circle is $K$.

From the cyclic quadrilateral we have that $\angle DMB=45^{\circ }$, but we have also that $\angle OCB=45^{\circ }$, so the lines $OC$, $DM$ are parallel, hence $KZ\perp DM$ and, since $DM$ is a chord of the circle, $ZD=ZM$. (1)

The triangle $CDF$ is isosceles and $CO$ is bisector, so $CO\perp DF$, and this means that $ZE\parallel DF$. It follows that $DFEZ$ is an isosceles trapezium, so $DZ=FE$ (2), and from (1) and (2) we have that $ZM=FE$ ().

From the isosceles trapezium we also have that $\angle FEZ=\angle DZE$ (3). Since $ZK$ is an altitude in the isosceles triangle $DZM$, it will be also angle bisector, so $\angle DZE=\angle MZE$ (4).

From (3) and (4) we conclude that $\angle FEZ=\angle MZE$, so $FE\parallel ZM$ (). From () and () we get that $FZME$ is a parallelogram, which is the desired result.