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. Solution Copyright © 1996-2008 Alexander Bogomolny                           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°. Reference R. Leikin, Dividable Triangles — What Are They?, Mathematics Teacher, May 2001, pp. 392–398. Copyright © 1996-2008 Alexander Bogomolny

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