Slovenia — National Math Olympiad

Let $G$ be the intersection of the segments $AC$ and $BF$. The points $A, B, C$ and $F$ lie on the same circle, so $\angle AFB = \angle ACB$. The diagonal $AC$ bisects the angle $\angle DCB$, so $\angle ACB = \angle DCA$. This implies $\angle AFG = \angle DCA$ and $\angle GFD + \angle DCG = \pi$. We conclude that the points $C, D, F$ and $G$ lie on the same circle.

The points $A, E, C$ and $D$ are concyclic, so $\angle AEC = \pi - \angle ADC$ and $\angle CEB = \pi - \angle AEC = \angle ADC$. Since $CDFG$ is a cyclic quadrilateral, we have $\angle FGC = \pi - \angle FDC$ and $\angle FGA = \angle FDC$. So, $\angle BGC = \angle FGA = \angle FDC$. We have shown that $\angle CEB = \angle ADC$ and $\angle BGC = \angle ADC$, so $\angle CEB = \angle BGC$. This implies that $BCGE$ is a cyclic quadrilateral. So, $\angle BEG = \pi - \angle ACB$, or $\angle GEA = \angle ACB$. At the same time we have $\angle ACB = \angle ACD$ because $AC$ is the bisector of the angle $\angle DCB$. Since $AECD$ is a cyclic quadrilateral, we have $\angle ACD = \angle AED$. We conclude that $\angle GEA = \angle DEA$ and the points $E, G$ and $D$ are collinear. Hence, $AC, BF$ and $DE$ meet at a point.