62nd Ukrainian National Mathematical Olympiad, Third Round, Second Tour

Consider the point $K$, symmetric to the point $B$ with respect to point $C$ (fig. 13). Then $\Delta BEC=\Delta KCE$, and $\angle KEC=\angle BEC=\angle AEF$, so points $K,E,F$ lie on the same line. Then from the statement it follows, that Fig. 13 $$\angle FAD=\angle EBC=\angle EKC=\angle FKD,$$ so points $A,K,D,F$ are concyclic. Therefore $$\angle BFD=180^{\circ }-\angle AFD=\angle AKD.$$ As $\Delta ABC=\Delta AKC$, and $\angle AKC=\angle ABC$, the equality above is rewritten as $\angle BFD=\angle ABC=\angle FBD$. From this in $\Delta FBD$ it follows that $DB=DF$.