51st Ukrainian National Mathematical Olympiad, 4th Round

Denote by $G$ the second point of intersection of the circles and join $G$ with all the vertices of the triangle $ABC$ (fig. 42). Then we have: $\angle GBC=\angle BAG=\alpha$ (the angle subtended by $BG$ of the first circle). On the other hand, $\angle GCD=\angle BAG=\alpha$ (the angle subtended by $DG$ of the second circle). Similarly, $\angle GBF=\angle CAG=\angle GCB=\beta$. So, $\angle FBC=\angle FCB=\alpha +\beta$, which implies that the triangle $BCF$ is isosceles ($BF=CF$).