1. if$\triangle ABC$ is equilateral$\implies$ $AE \perp BF$ $\triangle BDE$ is similar to $\triangle AMC$ wich$M$is midpoint of $\overline {AC}$ then $ \triangle BFE $ is similar to $ \triangle AME $ so $\angle MAE=\angle EBF$ then $ABMK$ is cyclic (wich $K$ is Intersection $AE$ of $BF$) then $AE \perp BF$ 2.$AE \perp BF$$\implies$ $\triangle ABC$ is equilateral $ABMK$ is cyclic then $\angle MAE=\angle EBF$ so $ \triangle BFE $ is similar to $ \triangle AME $ then $\triangle BDE$ is similar to $\triangle AMC$ then $ABMD$ is cyclic then $BD$ is bisector and altitude hence $\overline {BC}=\overline {AB}$ so$\overline {AB}=\overline {AC}=\overline {BC}$