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

ABC is an isosceles triangle with vertex angle

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Copyright © 1996-2017 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

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

So indeed ∠ACE = 10° and ∠AEC = 150°.

### Reference

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