## 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

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Copyright © 1996-2017 Alexander Bogomolny

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

Let F on AB be such that DF||BC and choose E on AB so that

On the other hand, ∠BDC = 180° - 40° - 80° = 60°. Therefore,

ΔBFD is also isosceles because ∠BDF = 180° - 80° - 60° = 40°, such that

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

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

There is a solution.

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Copyright © 1996-2017 Alexander Bogomolny

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