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 OB and CD respectively ,then $\mathrm{\angle }ANM=?$
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Jaelyn Levine
Step 1

Notice that $\mathrm{△}OAM\sim \mathrm{△}DAN$, $\mathrm{\angle }OAM=\mathrm{\angle }DAN$.
So, $\mathrm{\angle }NAM={45}^{\circ }$.
Step 2
Drop a perp from N to diagonal BD.
Notice that $\mathrm{△}TMN\cong \mathrm{△}OAM$
So, $AM=MN$ leading to $\mathrm{\angle }ANM={45}^{\circ }$.
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Ilnaus5
Step 1
$\phantom{\rule{thickmathspace}{0ex}}\mathrm{△}ANJ\phantom{\rule{thinmathspace}{0ex}}$ is an isosceles right triangle.

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 ${45}^{\circ }$.
Step 2
Let $\phantom{\rule{thinmathspace}{0ex}}AB=1\phantom{\rule{thinmathspace}{0ex}}$, $\phantom{\rule{thinmathspace}{0ex}}DN=z\in \left[0,\frac{1}{2}\right]\phantom{\rule{thinmathspace}{0ex}}$ and $\phantom{\rule{thinmathspace}{0ex}}\frac{OM}{OB}=\frac{DN}{DC}\phantom{\rule{thinmathspace}{0ex}}$ so that $\phantom{\rule{thinmathspace}{0ex}}z=\frac{1}{2}\phantom{\rule{thinmathspace}{0ex}}$ corresponds to the original diagram and $\phantom{\rule{thinmathspace}{0ex}}z=0\phantom{\rule{thinmathspace}{0ex}}$ corresponds to $\phantom{\rule{thinmathspace}{0ex}}M↦O,N↦D\phantom{\rule{thinmathspace}{0ex}}$. Then $\phantom{\rule{thinmathspace}{0ex}}OM=\frac{OB}{DC}\phantom{\rule{thinmathspace}{0ex}}DN=\frac{z}{\sqrt{2}}$, and:
$\mathrm{tan}\left(\mathrm{\angle }AND\right)=\frac{1}{z}\phantom{\rule{0ex}{0ex}}\mathrm{tan}\left(\mathrm{\angle }MNC\right)=\frac{1+z}{1-z}=-\phantom{\rule{thinmathspace}{0ex}}\frac{1+\frac{1}{z}}{1-\frac{1}{z}}=-\phantom{\rule{thinmathspace}{0ex}}\frac{\mathrm{tan}\left(\frac{\pi }{4}\right)+\mathrm{tan}\left(\mathrm{\angle }AND\right)}{1-\mathrm{tan}\left(\frac{\pi }{4}\right)\phantom{\rule{thinmathspace}{0ex}}\mathrm{tan}\left(\mathrm{\angle }AND\right)}\phantom{\rule{0ex}{0ex}}=\mathrm{tan}\left(\pi -\left(\frac{\pi }{4}+\mathrm{\angle }AND\right)\right)$
It follows that $\phantom{\rule{thinmathspace}{0ex}}\mathrm{\angle }MNC+\frac{\pi }{4}+\mathrm{\angle }AND=\pi \phantom{\rule{thinmathspace}{0ex}}$, and therefore $\phantom{\rule{thinmathspace}{0ex}}x=\frac{\pi }{4}\phantom{\rule{thinmathspace}{0ex}}$ regardless of z.
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