$O$ 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.
Best Answer
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 $\angle ANM = 45^\circ$.