Let ABC be an isosceles triangle (AB = AC) with BAC = 20°. Point D is on side AC such that CBD = 60°. Point E is on side AB such that BCE = 70°. Find the measure of CED.

Solution

Copyright © 1996-2008 Alexander Bogomolny

The solution is by Philippe Fondanaiche.

Find F on AB with BCF = 50°. This brings us into the classical framework so that we, for example, know that BDF = 30°.

In ΔBDF, DBF = 20° and BFD = 130°. In ΔCFE, ECF = 20°, CEF = 30° such that the two triangles are similar:

Since BFC = 50°, ΔBCF is isosceles and BF = BC. Also, DFE = 50°. With (1), this tells us that triangles BCF and DFE are similar, so that the latter is also isosceles and DF = DE. Hence, DEF = 50° and

CED = DEF - CEF = 50° - 30° = 20°.

Copyright © 1996-2008 Alexander Bogomolny