# The 80-80-20 Triangle Problem, Solution #6

Let ABC be an isosceles triangle

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Copyright © 1996-2018 Alexander Bogomolny

This is Solution 5 from [Knop] and is due to Aleksey Borodin, a high school student at the time.

Let O be the circumcenter of ΔBDE.

The inscribed angle ∠DBE = 30° and is subtended by the same arc as the central angle DOE. Thus

Triangles BDO and BCD are both isosceles implying that CO bisects ∠BCD. Thus

Now, in triangles CDE and COE, EO = ED, CE is a common side and

### 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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