What's the measure of the $\angle BAC$ in the question below?

Question

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

Answer 1

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 $\theta\approx 37^\circ$

Answer 2

Extend side $AB$ to point $X$ such that $CX\perp AX$. Since $\angle ABC=135^{\circ}$, $$\angle XBC=\angle XCB=45^{\circ}\implies BX=CX.$$ You have shown that $BC=15\sqrt{2}\;$ (using $\triangle ABC\sim \triangle FBC$). Hence, $$BX=CX=15.$$ In $\triangle AXC$, $$\sin \angle XAC=\frac{XC}{AC}$$ $$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\implies\angle XAC=\sin^{-1}\frac{15}{25}=\sin^{-1}\frac 35$$ $$\therefore\; \angle BAC\approx 37^{\circ} $$