Octarelate

Triangle $ABC$ and three others identical to it are pictured in red are organized to form a square as shown.

The original image made the triangle appear to be isosceles, but the problem did not state that it was and it does not need to be. So I make it a general right triangle.

The condition that $AM^2+NC^2=MN^2$ translates, thanks to the Pythagorean theorem, into $MN=NM’$ where $M'$ is a point equivalent to $M$ in the next triangle.

This means that the black octagon has sides of equal lengths.

It is inscribed into a square, so it could be tempting to assume that it is a regular octagon and that the angle

$\angle MON=\frac{360^\circ}{8}=45^\circ$

...

However, the octagon need not be regular, as shown in the lower image.

The triangles $ONM$ and $ONM’$ are no longer isosceles, but they are still congruent.

All of their sides are the same size, since the side $ON$ is shared, $MN=NM’$ and $OM=OM’$ .

Therefore the angles $\angle MON$ and $\angle NOM’$ are equal to each other.

They are also equal to the remaining 6 such angles.

So we can conclude that even in the general case

$\angle MON=\frac{360^\circ}{8}=\boxed{45^\circ}$

It doesn't specify where M and N are located. So let M be located at A. Then N bisects AC and BN bisects angle ABC. When you bisect a right angle, each of the two angles has a measurement of 90/2 = 45 degrees.