# Find an angle in an isosceles triangle.

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.

• Where is the question? – vitamin d Feb 27 at 3:48

## 3 Answers

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.

• Wow, this is genius. Thank you for this answer. – Student1058 Feb 27 at 5:56

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 $$180°$$. $$4 \alpha + 108°=180° \implies \alpha=18°$$

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 :) • I am asking about $\angle{AB'C}$ instead of $\angle{ABC}$. – Student1058 Feb 27 at 4:21
• Turned it into a full answer @Student1058. – dodoturkoz Feb 27 at 5:22

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. 