The 80-80-20 Triangle Problem, Solution #4

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

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

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

The 80-80-20 Triangle Problem

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