For reference (exact copy of question): In $\triangle ABC$, $\angle B $ measures $135^{\circ}$.
The cevian $ BF$ is traced so that $AF = 7 $ and $FC = 18.$
Calculate $\angle BAC$, if $\angle BAC = \angle FBC$. (answer:$37^{\circ} $)
My progress:
Here is the drawing I made according to the statement and the relationships I found
$\triangle ABC \sim \triangle FBC:\\
\frac{BC}{AC}=\frac{FB}{AB}=\frac{FB}{BC}\\
\frac{BC}{25}=\frac{FB}{AB}=\frac{18}{BC}\implies BC = 15\sqrt2$
it seems to me that the path is by auxiliary lines
Best Answer
Draw altitude from $B$ to $AC$ and let the foot be $D$. Say $DF=DB=x$. Now applying Pythagorean theorem to $\triangle BCD$ $$(18+x)^2+x^2=(15\sqrt 2)^2\implies x=3$$ So $BD=3$ and $AD=7-3=4$. It can be easily seen that $\triangle BDA$ is a $3:4:5$ triangle and hence $\theta\approx 37^\circ$