The 80-80-20 Triangle Problem, 60-70 Variant, Solution #2

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

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

∠FBO = ∠FCO = 20°,

quadrilateral BCOF is cyclic. In particular,

∠OFC = ∠ OBC = 60°.

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°.

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