67th Czech and Slovak Mathematical Olympiad

Let $O$ be the circumcenter of triangle $AEF$ and denote $\alpha =\angle BAC$. Since $\angle BAD$ and $\angle CAD$ are acute (Fig. 2), points $E$, $F$ lie in the half-plane $BCA$ and the Inscribed angle theorem yields $$\angle BED=2\cdot \angle BAD=\alpha =2\cdot \angle DAC=\angle DFC.$$



Fig. 2

The isosceles triangles $BED$ and $DFC$ are thus similar and we easily compute that $\angle EDF=\alpha$ and that $BC$ is the external $D$-angle bisector in triangle $DEF$.

Point $O$ lies on $BC$ and on the perpendicular bisector of $EF$. Framed with respect to triangle $DEF$, it lies on the external $D$-angle bisector and on the perpendicular bisector of the opposite side $EF$. Thus it is the midpoint of arc $EDF$ and $\angle EOF=\angle EDF=\alpha$. Quadrilateral $AEDF$ is a kite, hence $\angle EAF=\alpha$. Moreover, line $EF$ separates points $A$ and $O$, thus the Inscribed angle theorem implies that the size of the reflex angle $EOF$ is twice the size of the convex angle $EAF$. This yields $360^{\circ }-\alpha =2\cdot \alpha$ and $\alpha =120^{\circ }$.