Let $M$ be the midpoint of the base $AC$ of isosceles triangle $ABC$. Points $E$ and $F$ are marked on sides $AB$ and $BC$ respectively, such that $AE \neq CF$ and

$\angle FMC = \angle MEF = \alpha$. Find $\angle AEM$.
Consider the circumcircle of triangle $M E F$. The angle between the tangent and the chord $M F$ is equal to $\angle M E F = \angle F M C$. Therefore, $M C$ is the tangent. This means that the center of the circle lies on the altitude $B M$. Consequently, this altitude is the axis of symmetry of the figure. Since $A E \neq C F$, the circle intersects each of the lateral sides at two points. Moreover, $E$ and $F$ are not symmetric. The two possible cases are labeled accordingly (see figure).

![](https://cdn.mathpix.com/cropped/2024_05_06_7adbd1f2e5e559c8684fg-24.jpg?height=386&width=366&top_left_y=247&top_left_x=846)

1) The external angle $A E_{1} M$ of the inscribed quadrilateral $M E_{1} E_{2} F_{2}$ is equal to the angle $M F_{2} E_{2}$, and the latter is equal to the symmetric angle $M E_{1} F_{1}$, which is $\alpha$.
2) The inscribed angles $A E_{2} M$ and $M E_{2} F_{2}$, both equal to $\alpha$, subtend symmetric arcs.

## Answer

$\alpha$.