$O$ is intersection of diagonals of the square $ABCD$. If $M$ and $N$ are midpoints of $OB$ and $CD$ respectively ,then $\angle ANM=?$
Notice that $\triangle OAM \sim \triangle DAN$, $\angle OAM = \angle DAN$.
So, $\angle NAM = 45^\circ$.
Drop a perp from $N$ to diagonal $BD$.
Notice that $\triangle TMN \cong \triangle OAM$
So, $AM = MN$ leading to $\angle ANM = 45^\circ$.
Let a side of the the square be 1 unit.
Let the circle ADN cut BD at Z and AC at Z'.
Note that DN = NC; $AC = BD = \sqrt 2$ and by symmetry, BZ = CZ' = x, say.
By power of a point, $x \times \sqrt 2 = 1 \times 0.5$. This means $ x = \dfrac {\sqrt 2}{4}$.
That is Z is your M and hence, $\angle ANM = \angle ADB = 45^0$