The 80-80-20 Triangle Problem, Solution #1
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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.
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Solution
Copyright © 1996-2008 Alexander Bogomolny
Draw ED'||BC, D' on AC. Let BD' meet CE in P. Since ΔBCP is equilateral, CP = BC. Since ΔBCD is isosceles, BC = CD. Hence ΔCDP is isosceles,
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CPD = 80°,
DPD' = 40°.
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Since DD'P = 40°, ΔDPD' is isosceles and DP = DD'. Also, since ΔEPD' is equilateral, EP = ED'. Hence, by SSS, ΔDED' = ΔDEP, DE bisects D'EP so that CED = PED = 30°.
Reference
- H. S. M. Coxeter, S. L. Greitzer, Geometry Revisited, MAA, 1967, p. 159
- C. Knop, Nine Solutions to One Problem, Kvant, 1993, no 6, Solution 1.
Copyright © 1996-2008 Alexander Bogomolny
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