Belarusian Mathematical Olympiad

1. It is enough to prove the equalities $FM=BM=MC$ from which it will follow that $M$ is the midpoint of the hypotenuse of the right triangle $CFB$ and in particular $\angle CFB=90^{\circ }$.



Since the quadrilateral $ABDC$ is cyclic, it follows that $\angle ABC=\angle ADC$. Since points $M$, $D$, $E$ and $C$ lie on the same circle, $\angle MDC=\angle MEC$. From the cyclic quadrilateral $AFME$ we obtain the equalities $\angle BFM=\angle MEA=\angle FBM$. Therefore the triangle $BFM$ is isosceles with $FM=BM$ and $FM=BM=MC$.