## The 80-80-20 Triangle Problem, A Derivative, Solution #4

ABC is an isosceles triangle with vertex angle

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Copyright © 1996-2017 Alexander Bogomolny

This is a trigonometric solution reported in [Leikin].

Let E' be a point on AB with

By the Law of Sines, in ΔACE',

AE' / sin 10° = CE' / sin 20°.

But, by the double angle formula,

CE' = 2·AE'·cos 10°.

In ΔBCE',

BC / sin 30° = CE' / sin 80° = CE' / cos 10°,

so that

CE' = 2·BC·cos 10°,

implying AE' = BC and, subsequently, E = E'. Thus

### Reference

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