Let $P$ be such that $ADMP$ is a rectangle. Choose points $Q$ and $R$ on the line $AP$ such that $Q B D A$ and $A D C R$ are rectangles. Points $Q$, $B$ and $D$ lie on the circle of diameter $AB$, hence $ADEQ$ is a cyclic quadrilateral. Similarly, $R$, $C$ and $D$ lie on the circle of diameter $AC$, hence $ADFR$ is a cyclic quadrilateral.

The two quadrilaterals share a side, and have the same supporting lines for the other two sides. Since they are cyclic, the remaining two sides $EQ$ and $RF$ must be parallel. Thus $E$, $Q$, $R$ and $F$ are vertices of a trapezoid.

On the other hand, in rectangle $Q B C R$, $M$ is the midpoint of $BC$, and $MP$ is parallel to $QB$, so $P$ is the midpoint of $QR$. Since $N$ is the midpoint of $EF$, we obtain that, in trapezoid $Q E F R$, $NP$ is parallel to $QE$.

This implies that quadrilateral $ADNP$ is cyclic, having the sides parallel to the sides of $ADFR$. Moreover, $A$ lies on the circle circumscribed to this quadrilateral, because the other three vertices of the rectangle $ADMP$ lie on it. Hence the quadrilateral $ADMN$ is cyclic.

Consequently, $\angle ANM = 180^{\circ} - \angle ADM = 90^{\circ}$.