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 © 1996-2008 Alexander Bogomolny

This is Solution 3 from [Knop].

Draw a line through E parallel to BC and another through C parallel to AB. The two intersect in, say, H forming a parallelogram BCHE. Let P be the point on CE that completes the equilateral triangle BCP.

Then BP = BC = CD and BE = CH. EBP = 20° and also HCD = 20°; so that, by SAS, ΔEBP = ΔHCD. It follows that

CHD = BEC = 40°.

And, since CHE = 80°, HD is the bisector of angle CHE. On the other hand, CD is the bisector of angle ECH. The point D is, therefore, the incenter of ΔECH, the intersection of its angle bisectors. So DE is the bisector of angle CEH and CED = 30°.

Reference

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

Copyright © 1996-2008 Alexander Bogomolny