The 80-80-20 Triangle Problem, Solution #9
Let ABC be an isosceles triangle
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Copyright © 1996-2018 Alexander Bogomolny
This is Solution 8 from [Knop] and is due to Sergey Saprikin, a high school student at the time.
Let T be the intersection of the bisector of angle ACB with the side AB.
Then ∠BTC = 60°, and, since
In ΔCDT, E is the intersection of the angle bisector at C and of the external angle bisector T. T is then one of the excenters of ΔCDT. It follows that, in that triangle, DE is the external bisector of the angle at D.
Reference
- C. Knop, Nine Solutions to One Problem, Kvant, 1993, no 6.
The 80-80-20 Triangle Problem
- Solution #1
- Solution #2
- Solution #3
- Solution #4
- Solution #5
- Solution #6
- Solution #7
- Solution #8
- Solution #9
- Solution #10
- Solution #11
- Solution #12
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Copyright © 1996-2018 Alexander Bogomolny
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