Let ABC be an isosceles triangle (AB = AC) with BAC = 20°. Point D is on side AC such that CBD = 50°. Point E is on side AB such that BCE = 60°. Find the measure of CED.

Solution

Copyright © 1996-2009 Alexander Bogomolny

This is Solution 7 from [Knop] and is due to Alexander Kornienko, a high school student at the time.

Reflect B in AC and C in AB to obtain B' and, respectively, C'.

Now, AB'D = ABD = 30°. Further, B'AC' = 60° and AB' = AC' so it follows that B'D is the perpendicular bisector of AC'.

Since AC'E = ACE = CAE = C'AE, we see that E is equidistant from A, C, and C'. E then lies on the perpendicular bisector of AC', i.e. on B'D. But then

CED = 180° - 40° - 40° - 70° = 30°.

(This solution is neatly embedded into a configuration of an 18-gon.)

Reference

1. C. Knop, Nine Solutions to One Problem, Kvant, 1993, no 6.

Copyright © 1996-2009 Alexander Bogomolny