The 80-80-20 Triangle Problem, Solution #8

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 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

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

The 80-80-20 Triangle Problem

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