The 80-80-20 Triangle Problem, Solution #1

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

|Contact| |Front page| |Contents| |Geometry| |Up| |Store|

Copyright © 1996-2017 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,

∠CPD = 80°,
∠DPD' = 40°.

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

  1. H. S. M. Coxeter, S. L. Greitzer, Geometry Revisited, MAA, 1967, p. 159
  2. C. Knop, Nine Solutions to One Problem, Kvant, 1993, no 6, Solution 1.

The 80-80-20 Triangle Problem

|Contact| |Front page| |Contents| |Geometry| |Up| |Store|

Copyright © 1996-2017 Alexander Bogomolny

 61246268

Search by google: