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

ABC is an isosceles triangle with vertex angle ∠BAC = 20° and AB = AC. Point E is on AB such that AE = BC. Find the measure of ∠AEC.

Solution

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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 ∠ACE' = 10°. We'll show that AE' = BC and, therefore, E = E'.

By the Law of Sines, in ΔACE',

AE' / sin 10° = CE' / sin 20°.

But, by the double angle formula, sin 20° = 2·sin 10° · cos 10°, so that

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 ∠ACE = 10° and ∠AEC = 180° - 20° - 10° = 150°.

Reference

  1. R. Leikin, Dividable Triangles - What Are They?, Mathematics Teacher, May 2001, pp. 392-398.

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Copyright © 1996-2017 Alexander Bogomolny

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