Let $H$ and $O$ be the orthocenter and circumcenter of $\Delta ABC$ It's well-known that $AH$ is the isogonal conjugate of $AO$ and $AH=2AO.cos\hat{A}=AO$ if and only if $\hat{A}=60^{\circ}$ so $\Delta AOM\cong \Delta AHF$ and done

Solved with AI216, v4913, CT17, CyclicISLclesTrapezoid, kvedula2004 at 4. We have that $AF = AM = \frac{1}{2} AC$. However, $AFC$ is a right triangle, so we must have $$\cos\left(\angle BAC\right) = \frac{1}{2} \Rightarrow \angle BAC = 60^{\circ}$$

Draw segment $FM$. A well-known property says that $AM=MC=MF$; in particular, $AM=AF$ and triangle $AMF$ is isosceles. If $\angle BAC=60^\circ$, the above implies $\angle FAM=\angle AFM=60^\circ$ and thus triangle $AMF$ is equilateral. In particular, $AM=AF$. Conversely, if $AM=AF$, then $AMF$ is already an equilateral triangle, and in particular $\angle FAM=\angle BAC=60^\circ$.

The below mentioned famous theorem in right triangles is hidden: In any right triangle $FAC$ with $\angle CFA=90^o$ : $AC=2AF$ iff $\angle ACF =30^o$