$O$ is intersection of diagonals of the square $ABCD$. If $M$ and $N$ are midpoints of $OB$ and $CD$ respectively ,then $\angle ANM=?$ $O$ is intersection of diagonals of the square $ABCD$.  If $M$ and $N$ are midpoints of the segments $OB$ and $CD$ respectively, find the value of $\angle ANM$.

Here is my approach:
Assuming the length of the square is $a$. We have $\tan(\angle AND)=2$ and I draw a perpendicular segment from $M$ to $NC$ and calling the intersection point $H$ then $\tan (\angle MNH)=\dfrac{\frac34a}{\frac a4}=3$ ($MH$ can be found by Thales Theorem in $\triangle BDC$)
Hence
$$\angle ANM=180^{\circ}-(\tan^{-1}2+\tan^{-1}3)=45^{\circ}$$

I'm looking for other approaches to solve this problem if it is possible.
Intuitively, If I drag the point $N$ to $D$ and $M$ to $O$ (the angle is clearly $45^{\circ}$ here) then by moving $N$ from $D$ to $C$ and $M$ from $O$ to $B$ with constant speed, I think the angle remain $45^{\circ}$. But I don't know how to prove it.


Note that $\small \triangle ADN$ is a right triangle with perpendicular sides in the ratio $1:2$.
Draw $\small NT$ parallel to $\small CA$.
From $\small \triangle DOC$, $\small DT=TO$ (midpoint theorem).
Thus, can you see $\small TM=2\ TN$?
Therefore $\small \triangle MTN$ is also a right triangle with perpendicular sides in the ratio $1:2$.
$\small \therefore\triangle ADN\sim\triangle MTN\implies\angle DAN=\angle TMN\implies ADNM\text{ is cyclic}$
$\small \implies x= \angle ANM=\angle ADM=45^\circ$
