## The 80-80-20 Triangle ProblemA Curious PartitionInverse Problem

In ΔABC, ∠B = 80°. The bisector of angle B meets AC in D such that AD = BC + BD. Find angle C.

Solution

In ΔABC, ∠B = 80°. The bisector of angle B meets AC in D such that AD = BC + BD. Find angle C.

Let F on AB be such that DF||BC. Then

∠BDF = ∠CBD = ∠DBF = 40°,

making, in particular, ΔBDF isosceles:

BF = DF = CD.

Next, since DF||BC, ∠ADF = ∠ACB = ∠DCB. Let E be on AD such that DE = BC. Triangles BCD and EDF are equal by SAS so that EF = BD.

But we know that AD = BC + BD and DE = BC. This yields AE = BD so that ΔAEF is isosceles: AE = EF = BD.

∠DEF = ∠CBD = 40°, implying ∠FAE = ∠EAF = 20°. Finally, in ΔABC, ∠ABC = 80° and ∠BAC = 20°, leaving ∠ACB = 80°.

This problem is an inverse to another one related to the 80-80-20 triangle.

Note: the above solution has been observed to contain a logical flaw; an implicit assumption made in the proof happened to be equivalent to the statement of the problem. The corrected solution appears elsewhere.

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