Find an angle in an isosceles triangle.

This answer uses a somewhat miraculous construction, so it may be less direct than the "direct angle chase" suggested by the other answer.

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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.


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 :)

enter image description here

Tags:

Geometry

Angle

Euclidean Geometry

Triangles

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