$O$ 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.

Hint: $\;\triangle ANJ\,$ is an isosceles right triangle.


[ EDIT ] The following is about the second part of OP's question.

Intuitively, If I drag the point $N$ to $D$ and $M$ to $O$  [...]  then by moving $N$ from $D$ to $C$ and $M$ from $O$ to $B$ with constant speed, I think the angle remain $45^{\circ}$.

Let $\,AB=1\,$, $\,DN=z \in \left[0, \frac{1}{2}\right]\,$ and $\,\frac{OM}{OB}=\frac{DN}{DC}\,$ so that $\,z=\frac{1}{2}\,$ corresponds to the original diagram and $\,z=0\,$ corresponds to $\,M \mapsto O, N \mapsto D\,$. Then $\,OM = \frac{OB}{DC} \,DN = \frac{z}{\sqrt{2}}$, and:
$$
\tan\left(\angle AND\right) = \frac{1}{z}
\\ \tan\left(\angle MNC\right)=\frac{1+z}{1-z} = -\,\frac{1+\frac{1}{z}}{1 - \frac{1}{z}} = -\,\frac{\tan\left(\frac{\pi}{4}\right)+\tan\left(\angle AND\right)}{1-\tan\left(\frac{\pi}{4}\right)\,\tan\left(\angle AND\right)} \\ = \tan\left(\pi -\left(\frac{\pi}{4}+\angle AND\right)\right)
$$
It follows that $\,\angle MNC + \frac{\pi}{4} + \angle AND = \pi\,$, and therefore $\,x = \frac{\pi}{4}\,$ regardless of $\,z\,$.
