The 80-80-20 Triangle Problem, 60-70 Variant, Solution #2
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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.
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Solution
Copyright © 1996-2008 Alexander Bogomolny
The solution is by Radu Ionescu.
Find F on AB with BCF = 50°. This brings us into the classical framework so that we, for example, know that CFD = 80°.
Let O be the intersection of BD and CE and H the intersection of BD and CF. Since
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FBO = FCO = 20°,
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quadrilateral BCOF is cyclic. In particular,
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OFC = OBC = 60°.
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From the classical problem, CFD = 80° which implies that OFD = 20°.
In ΔDFH, HFD = (CFD) = 80°. FHD = 70°. Thus, FDH = 30°. Since FDO is the same angle, FDO = 30°. On the other hand, FEO = BEC = 30° also. So that the quadrilateral FODE is cyclic making CED = OED = OFD = 20°.
Copyright © 1996-2008 Alexander Bogomolny
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