How can I prove that the cyan triangle in the picture is isosceles?

Question

There is a circumference $\gamma$ which has a square $ABCD$ inscribed. There is also a semi-circumference $\gamma_1$ centered in $H$ with radius $\overline{OH}=\overline{AH}=\overline{HB}$. Then there is a half line having origin in $A$ and intersecting $\gamma_1$ and $\gamma$ in $E$ and $F$ respectively. I need to prove that the cyan triangle $BEF$ is right-angled and isosceles using only euclidean geometry theorems (except triangle similitude criteria). I managed to prove that the angle $B\hat{E}F$ is right, then I tried to extend the segment $\overline{BE}$ intersecting $\gamma$ in G and tried to prove that $GEF \equiv ABE$ with one of the congruence criteria, but no luck. I then tried a lot of constructions, for example drawing the semi-circumference intersecting the points $E$, $F$, and $B$, but I still get stuck on proving that there is some length on one side which is equal to some other length on the other side.

Answer 1

Hint: $\angle ACB = \angle AFB = 45^\circ$ follows from angles in the same segment.

Answer 2

First, note that $\angle ACB$ and $\angle AFB$ subtend the same arc. Then, since $\angle ACB = 45^{\circ}$, $\angle AFB = 45^{\circ}$. Also, $\angle AEB$ is inscribed in a semicircle, so it is a right triangle. Then, $\angle FEB = 90^{\circ}$ and $\angle EFB = 45^{\circ}$, so $\angle FBE = \angle EFB = 45^{\circ}$, and $\triangle FBE$ is isosceles. $\blacksquare$

Answer 3

Note that

$$2\angle AFB= \angle AOB =\angle AEB =90^\circ$$ which leads to $\angle AFB = \angle FBE =45^\circ$.