The 80-80-20 Triangle Problam, Solution #2

The 80-80-20 Triangle Problem,
A Curious Partition, Solution

The 80-80-20 triangle can be partitioned in a curious way: let BD, with D on AC, bisect angle B. Then AD = BC + BD.

The 80-80-20 triangle can be partitioned in a curious way: let BD, with D on AC, bisect angle B. Then AD = BC + BD.

Let F on AB be such that DF||BC and choose E on AB so that BD = DE. ΔBDE is isosceles, with ∠EBD = 40°, by construction. This implies that ∠BED = 40° and hence ∠BDE = 100°.

On the other hand, ∠BDC = 180° - 40° - 80° = 60°. Therefore, ∠ADE = 180° - 100° - 60° = 20°, implying that ΔAED is isosceles with AE = DE.

ΔBFD is also isosceles because ∠BDF = 180° - 80° - 60° = 40°, such that (CD =) BF = DF.

Finally, ∠EDF = 80° - 20° = 60° = ∠EDC. We see that triangles BCD and EFD are equal (by SAS), implying BC = EF.

It follows that

	AD	= AF
		= AE + EF
		= BD + BC.

As a follow up, an inverse problem has been posted at the old CTK Exchange:

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

There is a solution.

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The 80-80-20 Triangle Problem,A Curious Partition, Solution

The 80-80-20 Triangle Problem,
A Curious Partition, Solution