euclidean-geometrygeometryplane-geometry
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:
$\triangle ABF: cos(\frac{C}{2}) = \frac{x}{AF}\\
AF^2 = x^2+BF^2\\
\triangle AFC: Law ~of~ cosines:\\
AF^2 = x^2+AC^2-2.x.AC.cosC\\
\triangle ABC:\\
cos C = \frac{BC}{AC} =\frac{x+BC}{AC}\\
x^2+(x+BF)^2 = AC^2\\
Th.Stewart \triangle ABC:\\
AC^2.BF+x^3=AF^2BC+BC.x.BF$
…??
Observe, $$\angle ICE=\angle IBE=90^{\circ}\implies BECI \;\text{is cyclic}. $$ Let $M$ be the midpoint of $IE$ and the center of the circumcircle of quadrilateral $BECI$. We have, $$ IM=EM=CM=AC.$$ Hence, $$\angle CAM=\angle CMI=2\angle CBI=16^{\circ}$$ Therefore, $$\angle ACB=180^{\circ}-\angle A-\angle B= 180^{\circ}-32^{\circ}-16^{\circ}=\boxed{132^{\circ}}$$
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 $\theta\approx 37^\circ$
Best Answer
Here is a construction that makes things simple. Extend $CB$ such that $BE = BF = y$
Now $ \displaystyle \angle BAE = \frac{\angle C}{2} \implies \angle CAE = 90^0 - \frac{\angle C}{2}$
And we notice that $\triangle ACE$ is isosceles so $AC = x + 2y$
Applying Pythagoras in $\triangle ABC$,
$(x+2y)^2 = x^2 + (x+y)^2$
$4y^2 = x^2 + y^2 - 2xy = (x-y)^2$
That leads to $x = 3y$ and sides of $\triangle ABC$ are in the ratio $3:4:5$