Prove the independence of a certain segment in a special triangle Let $\triangle ABC$ be a triangle with obtuse angle $\angle A$ and $\overline{AB} = 1$. Also, let $\angle C = \gamma$ and $\angle B = 2\gamma$. If $E$ and $F$ are intersection points of perpendicular bisector of $\overline{BC}$ and circle $(A, \overline{AB})$, prove that $\overline{EF}$ is constant (not dependent of $\gamma$).
Note that circle $(A, r)$ means a circle with point $A$ as its center and $r$ as its radius.
It's not hard to see that $\angle FAE = 120^\circ$ but I don't know how to prove it either. Hope someone can help.
 A: Note that if we fix the length $AB$, then the length of the chord $EF$ depends only on the distance from $A$ to $\overline{EF}$. Let $\overline{AD}$ be an altitude of $\triangle ABC$. Then
\begin{align*}
BD &= AB \cos(2\gamma) & BC &= AD \cot(2\gamma) + AD \cot(\gamma) = AB \sin(2\gamma) (\cot(2\gamma) + \cot(\gamma))
\end{align*}
Note that the distance from $A$ to $\overline{EF}$ is nothing but
$$\frac{BC}2-BD = AB \left( \frac12 \sin(2\gamma) (\cot(2\gamma)+\cot(\gamma)) - \cos(2\gamma) \right)$$
We need to show, then, that the above is a constant function of $\gamma$. Writing things out with sines and cosines, we get
\begin{align*}
\frac{BC}2-BD &= AB \left( \frac12 \sin(2\gamma) \left( \frac{\cos(2\gamma)}{\sin(2\gamma)} + \frac{\cos(\gamma)}{\sin(\gamma)} \right) - \cos(2\gamma) \right) \\
&= AB \left( \frac12 \cos (2\gamma) + \frac 12 \cdot \frac{\cos(\gamma)\sin(2\gamma)}{\sin(\gamma)} - \cos(2\gamma) \right) \\
&= \frac{AB}2 \left( \frac{\cos(\gamma)\sin(2\gamma)}{\sin(\gamma)} - \cos(2\gamma) \right) \\
&= \frac{AB}2 \left( \frac{\cos(\gamma) \cdot 2 \sin(\gamma) \cos(\gamma)}{\sin (\gamma)} - (2\cos^2(\gamma)-1) \right) \\
&= \frac{AB}2 \left( 2\cos^2(\gamma) - (2\cos^2(\gamma)-1) \right) \\
&= \frac{AB}2
\end{align*}
Indeed, we see that this is independent of $\gamma$, which implies that $EF$ is independent of $\gamma$ as well, as mentioned above.
There is almost certainly a faster, more purely geometric way to prove the result, but this gets the job done.
A: Reflect $A$ across $\overline{EF}$ to $D$, creating isosceles trapezoid $\square ABCD$ and rhombus $\square AEDF$.

Note that, since we're given $\angle ABC=2\angle ACB$, we find $\overline{AC}$ bisects $\angle BCD$, so that we have
$$\angle DCA\;\cong\;\angle ACB \underbrace{\cong}_{\overline{AD}\,\parallel\,\overline{BC}} \angle CAD \quad\to\quad \overline{AD}\cong\overline{CD}\;\left(\;\cong\overline{AB}\cong\overline{AE}\;\right)$$
We conclude that $\overline{AD}$ divides the rhombus into two equilateral triangles, so that $|EF|=|AB|\sqrt{3}$, which is independent of the angles at $B$ and $C$. $\square$
