TITLE: What's the measure of the $\angle BAC$ in the question below?
QUESTION [3 upvotes]: 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

REPLY [2 votes]: 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 $\theta\approx 37^\circ$

REPLY [1 votes]: 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 $\triangle AXC$,
$$\sin \angle XAC=\frac{XC}{AC}$$ $$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\implies\angle XAC=\sin^{-1}\frac{15}{25}=\sin^{-1}\frac 35$$
$$\therefore\; \angle BAC\approx 37^{\circ} $$
{"set_name": "stack_exchange", "score": 3, "question_id": 4268377}