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.