Enough to prove XN is collinear with the midpoint of AB called O. But because AB is diameter AXB is 90 degrees and ONX collinear by Iran Lemma.

It's much easier: If we think of the right triangle $AXB$ as fixed, then as $C$ moves along the reflection of $AB$ over $BX$, the midpoint $N$ of course moves along a line parallel to $BC$ (the midline). But one such choice is $X$ (when $BC=AB$), hence the claim.

Let $M$ be midpoint of $BC$. Then it satisfies to prove $X, N, M$ collinear. Now angle chase $\angle AMX=2\angle ABX=\angle ABC=\angle AMN, so M, N, X collinear and done.