The 80-80-20 Triangle Problem, A Derivative, Solution #5

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.


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This solution has been reported in [Leikin]. It applies the same strategy as Solution #5, viz., assuming ∠ACE = 10° and subsequently proving that AE = BC.

Form ΔACS congruent to ΔABC:


Since ∠ACE = 10°, CE serves as an angle bisector in the isosceles ΔACS. Hence, it is also the altitude and the median from C. Let Z be its foot, the midpoint of AS. We have

  AZ = AS / 2 = BC / 2.

But in right ΔAEZ, ∠EAZ = 60° which makes AZ = AE /2. And we get the needed identity: AE = BC.

So indeed ∠ACE = 10° and ∠AEC = 150°.


  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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