MathNet 2003 — Part 02 Question 34

The distinct points $M$ and $N$ lie on the hypotenuse ($AC$) of the right isosceles triangle $ABC$ so that $M \in (AN)$ and $MN^{2} = AM^{2} + CN^{2}$. Prove that $m(\angle MBN) = 45^{\circ}$.