The core idea (and the only idea i guess) is that $K$ lies on the bisector (external) of $\angle{BAC}$. This is well-known but its proof is short so anyways... define $N'$ to be the second intersection of $(MKA)$. Thus, by the angles formed, we obtain $MK = N'K$ so we've proved that $N' = N$. Thus let $A'$ be the diametrically oposed to $A$ wrt. $(AMN)$. Hence $A'K \perp KA$ which means $K, E, A'$ are collinear but also $\angle{MA'K} = \angle{KA'N}$. So $A'D = A'F$ and the conclusion follows. P. S. This indeed shows that the fact of drawing the circle centered at $K$ with radius $KE$ is not needed, we just used $E$ to say that $KE$ is perpendicular to $BC$.

We have the following main Claim. Claim: Triangle $KMN$ is equilateral. Proof: Note that $2R_{AKM}=\dfrac{KM}{\sin \angle KAM}=\dfrac{KN}{\sin \angle KAN}=2R_{AKN},$ hence triangles $AKM$ and $AKN$ have the same circumradious, therefore $K \in (AMN),$ which implies that $\angle MKN=\angle MAN=60^\circ,$ as desired $\blacksquare$ Now, let $DM,FN$ intersect at point $S$. Then, $\angle AMS=\angle ANS=90^\circ,$ hence $S \in (AMN)$. Thus, $\angle AKS=90^\circ=\angle AKE,$ implying that points $K,E,S$ are collinear. Now the finish is trivial, as triangle $DFS$ is isosceles ($\angle KSD=\angle KSM=\angle KNM=\angle KMN=\angle KSN=\angle KSF$), and $SK$ is its $S-$ altitude.