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 6 from [Knop] and is due to Maria Gelband, a high school student at the time.

Let M be on the bisector of angle BCD with the property that ABM = 30°.

Since ΔBCD is isosceles (BC = CD), CM is the perpendicular bisector of BD so that, BM = DM. From the choice of M, DBM = 60° implying that ΔBDM is equilateral.

Since BE is the bisector of DBM it is also the perpendicular bisector of DM.

In ΔCDM, E is the intersection of the perpendicular bisector of DM and the bisector of the angle at C. Thus E belongs to the circumcenter of ΔCDM. In the circumcircle, two inscribed angles, CED and CMD are equal, but the latter is equal half of BMD making CED = 30°.

Reference

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

Copyright © 1996-2008 Alexander Bogomolny