Point $M$ is the midpoint of chord $A B$. Chord $C D$ intersects $A B$ at point $M$. A semicircle is constructed on segment $C D$ as its diameter. Point $E$ lies on this semicircle, and $M E$ is perpendicular to $C D$. Find the angle $A E B$.
$M E$ - the height of the right triangle $C E D$, drawn from the vertex of the right angle, so $E M^{2}=C M \cdot M D$ $=A M \cdot M B=A M^{2}$, that is, the median $E M$ of triangle $A E B$ is equal to half of the side $A B$. Therefore, $\angle A E B=$ $90^{\circ}$.

## Answer

$90^{\circ}$.