It can be noted that $ME \neq MD$ for $AE = AD$. We have, $AM$ is the external bisector of $\angle EAD$ and $ME = MD$. Thus $AMED$ is cyclic. Thus we have \[\angle BFM = 180^{\circ} - \angle FME - \angle MEF = 180^{\circ} - \angle DAE - \angle MEF = 60^{\circ} - \angle MEF = \angle BME.\]

Let $N,P$ the projections of $M$ onto $CA, AB$ respectively; as $MN=MP$, with $MD=ME$, we get $\triangle DMN\cong\triangle EMP$, which implies $\angle ADM=\angle AEM \ (\ 1\ )$, making $MADE$ cyclic, i.e. $\triangle ADF\sim\triangle MEF$, done. Best regards, sunken rock

Very easy by theorem 9 ($Mr.vakili$)

Let angle BFM =x angle DMC=60°+x, angle ACM =60° angle MDC=60°-x let angle BME=y, angle MBE=120° angle BEM=60°-y In ∆DMC, MD/ MC=sin 60°/ sin(60°-x) In ∆BME, ME/ MB=sin 60°/ sin(60°-y) Since, MB=MC MD=ME sin 60°/ sin(60°-x)=sin 60°/ sin(60°-y) sin(60°-x) =sin(60°-y) 60°-x=60°-y x=y So, angle BFM=angle BME @KRISHIJIVI

Let $G$ be a point on line $EG$ such that line $EG$ is parallel to line $BC$ and $MG=ME=MD$ ⟹ $M$ is the center of a circle with point $E,D,G$ on it. extend ray $EM$ to meet the circle at point $H$ (Not $E$) We have to prove $\angle BFM = \angle BME$ but $\angle BME = \angle MEG =\angle HEG =\angle HDG$ also $\angle MGD =\angle MDG =60-\angle MEG$ $\therefore \angle MDH = 60$ $\therefore$ since $\angle EDH =90, \angle EDM =30$ $\therefore \angle DME = 120$ $\therefore$ since $\angle MEF = 60-\angle MEG, \angle EFM= MEG, \therefore \angle BFM =\angle BME$ $Q.E.D$ I also thought this could be solved via barycentric coordinates but couldn't figure it out exactly