What's the measure of the $\angle BAC$ in the triangle below? For reference: In the right triangle $ABC$, right at $B$, the corner $AF$ is drawn such that $AB = FC$ and $\angle ACB = 2 \angle BAF$. Calculate $\angle BAC$.
My progress:

$\triangle ABF: cos(\frac{C}{2}) = \frac{x}{AF}\\
AF^2 = x^2+BF^2\\
\triangle AFC: Law ~of~ cosines:\\
AF^2 = x^2+AC^2-2.x.AC.cosC\\
\triangle ABC:\\
cos C = \frac{BC}{AC} =\frac{x+BC}{AC}\\
x^2+(x+BF)^2 = AC^2\\
Th.Stewart \triangle ABC:\\
AC^2.BF+x^3=AF^2BC+BC.x.BF$
...??

If $y=BF$ and $t=\tan\frac{C}{2}=\frac{y}{x}$, then $\tan C=\frac{x}{y+x}=\frac{1}{1+t}$
Also $\tan{C}=\frac{2t}{1-t^2}$ from the double angle formula. It follows that:
$$\frac{2t}{1-t^2}=\frac{1}{1+t}$$
And you cand find $t=\frac{1}{3}$ by solving this equation. It follows that $\tan C=\frac{3}{4}$, $\tan\angle BAC=\frac{4}{3}$, and finally $\angle BAC=\arctan\frac{4}{3}$
