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 © 1996-2009 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

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

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

Copyright © 1996-2009 Alexander Bogomolny