## Corrected

Let's start by drawing a figure.

!

First, since $AB < AC$, $D$ lies on the segment $[AC]$ and the angle $\angle MAE$ is obtuse. Since $MC = MA$, this implies that $\triangle CAM$ and $\triangle AME$ are similar if and only if $\triangle AME$ is isosceles at $A$.

Suppose that $\triangle AEM$ and $\triangle MCA$ are similar. From the first paragraph, this implies $AM = AE$. The lines $(EM)$ and $(AC)$ are parallel, so the lines $(AB)$ and $(ME)$ are perpendicular. Combined with the fact that $AM = AE$, this shows that $\triangle AMB$ and $\triangle ABE$ are symmetric with respect to $(AB)$. Let $\alpha = \angle MAB$. Since $M$ is the midpoint of $[BC]$, we have $AM = MB$, which gives $\alpha = \angle ABC$. On the other hand, $\alpha = \angle BAE = \angle ABE = 90^\circ - \angle DBA = \angle ADB$.

Since $\angle DAB = \angle DMB = 90^\circ$, the points $A, D, M, B$ are concyclic. We deduce that $\angle AMB = \angle ADB = \alpha$. The triangle $\triangle AMB$ has all three angles equal to $\alpha$. Therefore, it is equilateral and $\alpha = 60^\circ$. Thus, $\angle ABC = 60^\circ$.

Conversely, if $\angle ABC = 60^\circ$, the same reasoning shows that $\triangle AMB$ and $\triangle ABE$ are symmetric with respect to $(AB)$, which implies that $\triangle AME$ is isosceles at $A$. Since $\angle CAM = \angle AME$, $\triangle AEM$ and $\triangle MCA$ are similar.