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

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

• you are close given you already found $BC$. Drop a perp from $C$ to $AB$ extend. Oct 5, 2021 at 13:02
• @MathLover..Yeah..sometimes we try to find "miraculous" solutions but suddenly the simplest is the most efficient..thanls Oct 5, 2021 at 16:50

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 $$\theta\approx 37^\circ$$ 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 $$\triangle AXC$$, $$\sin \angle XAC=\frac{XC}{AC}$$ $$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\implies\angle XAC=\sin^{-1}\frac{15}{25}=\sin^{-1}\frac 35$$ $$\therefore\; \angle BAC\approx 37^{\circ}$$