Question
Given $\triangle ABB'$, where $\angle BAB'=108^{\circ}$.
$\overline{BC}$ is the bisector of $\angle ABB'$, and $\overline{AB}=\overline{AB'}=\overline{B'C}$. Find $\angle{AB'C}$.
I prefer solutions without trigonometric functions, but answers with them are also fine.
I have tried constructing a point $D$ on $\overline{AB}$ such that $\overline{AB}=\overline{BD}$. Therefore, both $\triangle{AB'D}$ and $\triangle{ACD}$ are isosceles. However, the problem remains unsolved after such construction. I can't see the correct auxiliary lines to make.
Any suggestions, hints, or even full solutions (you don't have to) are appreciated.
Thanks for reading my post.
Best Answer
This answer uses a somewhat miraculous construction, so it may be less direct than the "direct angle chase" suggested by the other answer.
Construct $D$ such that $D$ lies on $AB$ and $AB' = B'D$. Then $\angle ADB' = \angle B'AD = 72^\circ$.
As $\angle ABB' = 36^\circ$, $\triangle BDB'$ is isosceles as well. Hence $BE$ is the perpendicular bisector of $B'D$.
$C$ lies on $BE$, hence $B'C = CD$. As we are given $DB' = AB' = CB'$, $\triangle B'CD$ is equilateral, and thus $\angle AB'C = 24^\circ$ after some calculations.