NMO Selection Tests for the Junior Balkan Mathematical Olympiad

Let $G$ be the centroid of the given triangle. Since $DF\parallel AC$, one has $\angle DAC=\angle GDF$.

If $\angle DAC=\angle ABE$, then $\angle GDF=\angle FBG$, hence the quadrilateral $BFGD$ is cyclic, implying $\angle AFC=\angle BDA$. The converse holds by the same argument.