Solution:

$\angle AQ M = \angle EQ D$ (opposite angle) $= \angle EDC$ ($CD$ tangent to circle $EQD$) $= \left(180^\circ - \angle C\right) / 2 = \angle A / 2 + \angle B / 2$ $(*)$.

$MF^2 = MQ \cdot ME$ ($MF$ tangent to circle $FQE$). So $AM = AF$ is equivalent to $AM^2 = MQ \cdot ME$ or $AM / MQ = ME / AM$. But since triangles $AMQ$ and $EMA$ have a common angle $M$, $AM / MQ = ME / AM$ iff they are similar, and hence iff $\angle AQ M = \angle A$. Using $(*)$, $AM = AF$ iff $\angle A = \angle B$.