Argentine National Olympiad 2015

The conditions give $BF=BD$ ($=\frac{2}{3}AB$), also $D\hat{B}F=60^{\circ }$, hence triangle $DBF$ is equilateral. Then $$DF\parallel AC\text{ as }B\hat{D}F=B\hat{A}C=60^{\circ }.$$ Hence $C\hat{D}F=A\hat{C}D$.

On the other hand $A\hat{C}D=B\hat{C}E$ by the symmetry of the figure (or because triangles $ADC$ and $BEC$ are congruent). Then $C\hat{D}F=B\hat{C}E$. So



the required sum $C\hat{D}F+C\hat{E}F$ is equal to $F\hat{C}E+C\hat{E}F$. By the exterior angle theorem that last sum equals $B\hat{F}E$. Now $FE$ is a median in the equilateral triangle $DBF$, hence also a bisector. Therefore $B\hat{F}E=\frac{1}{2}B\hat{F}D=\frac{1}{2}\cdot 60^{\circ }=30^{\circ }$, which is the answer to the problem.