Here’s a sketch. (again, I’ll be writing detailed sol when I get my notebook back (which is tomorrow yay)) WLOG $\angle B\geq \angle C$ Easy angle chasing gives $FA=FC$ and $\angle AFC=90^{\circ}=2\cdot 45^{\circ}=2\cdot AXC$ so it’s the circumcenter of $AXC$. After this, just angle chase to get $\angle B=75^{\circ},\angle C=60^{\circ}$, and permutation.

Well, actually there is another solution $\beta=15^\circ$ and $\gamma=120^\circ$. The reason for this phenomenon is that what the angle chasing actually yields is that $3\beta=45^\circ$ with directed angles modulo $180^\circ$ so that there are several solutions. After all, a funny twist!