Find an angle in an isosceles triangle.

Question

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.

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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.

Answer 1

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.

Answer 2

I wanted you to notice it yourself, but I have already received a downvote so I will spoil the fun.

If $BC$ is an angle bisector and $\bigtriangleup ABB'$ an isosceles triangle can you find $\measuredangle ABC=\alpha$? Name the angles, say $\alpha$. The interior angles of $\bigtriangleup ABB'$ should sum to $180°$. $$4 \alpha + 108°=180° \implies \alpha=18°$$

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I was trying to show that this is a special triangle, in particular a golden gnomon. Therefore, the side $EC$ equals: $$EC=DC \cdot \phi$$

Now, apply the law of sines in $\bigtriangleup EFC$ to get $\measuredangle EFC=150°$.

It may help to recall that (which of often comes up in geometry problems, proof ): $$\sin(18°)=\frac{1}{2\phi}$$

The interior angles of $\bigtriangleup EFC'$ should sum to $180°$. So, $\measuredangle ECF=12°$.

We know that $\measuredangle DCE=36°$, so you can easily find $\measuredangle AB'C=24°$.

Just for fun I completed the diagram to a regular pentagon :)

Answer 3

A circle can be imagined with center at B' and radius AB'... etc. But sincere apologies, as...

after constructing a perpendicular to one side of regular pentagon I thought we obtain point C and $x^0 =24^0$ by direct angle chase alone.. although one consideration is still missing.