# The 80-80-20 Triangle Problem, Solution #9

Let ABC be an isosceles triangle

|Contact| |Front page| |Contents| |Geometry| |Up|

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

|Contact| |Front page| |Contents| |Geometry| |Up|

Copyright © 1996-2018 Alexander Bogomolny

65100782 |