<s>[INST] 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:

$\triangle ABF: cos(\frac{C}{2}) = \frac{x}{AF}\\
AF^2 = x^2+BF^2\\
\triangle AFC: Law ~of~ cosines:\\
AF^2 = x^2+AC^2-2.x.AC.cosC\\
\triangle ABC:\\
cos C = \frac{BC}{AC} =\frac{x+BC}{AC}\\
x^2+(x+BF)^2 = AC^2\\
Th.Stewart \triangle ABC:\\
AC^2.BF+x^3=AF^2BC+BC.x.BF$
...??
 [/INST] Just a note: This is a method for reference, it approximates your angle answer:
We know that:
$\tan(\frac{c}{2}) = \frac{BF}{x}$
$\tan(90-c) = \frac{x+BF}{x} = 1 + \frac{BF}{x} = 1+\tan(\frac{c}{2})$
Using the relation between sine, cosine and tangent:
$\frac{\cos(c)}{\sin(c)}$ = $1+\frac{\sin(\frac{c}{2})}{\cos(\frac{c}{2})}$
Then using the half-angle formulae:
$\frac{\cos(c)}{\sin(c)}$ = $1\pm\frac{\sqrt{\frac{1-\cos(c)}{2}}}{\sqrt{\frac{1+\cos(c)}{2}}}$
$\frac{\cos(c)}{\sin(c)}$ = $1+\frac{\sqrt{\frac{1-\cos^2(c)}{4}}}{\frac{1+\cos(c)}{2}}$
Using Pythagorean-Trig identity:
$\frac{\cos(c)}{\sin(c)}$ = 1+$\frac{\sin(c)}{\cos(c)+1}$
$\cos^2(c)+\cos(c)$=$\sin^2(c) +\sin(c)\cos(c)+\sin(c)$
$\cos(2c)+\cos(c)-\sin(c)\cos(c)-\sin(c)=0$
$2\cos(2c)+2\cos(c)-\sin(2c)-2\sin(c)=0$
Because $c$ is real:
$(\cos(\frac{c}{2}))(-\sin(c)+2\cos(c)-1)=0$
$\sin(c)=2\cos(c)-1$
$\sqrt{1-\cos^2(c)} = 2\cos(c)-1$
$4\cos^2(c)-4\cos(c)+1=1-\cos^2(c)$
$(\cos(c))(5\cos(c)-4) = 0$
--> $\cos(c)=\frac{4}{5}$, given that $0<c<90$
$c \approx 37^o$
$180^o - 37^o - 90^o = 53^o$
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