# The 80-80-20 Triangle Problem, Solution #1

Let ABC be an isosceles triangle

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Copyright © 1996-2018 Alexander Bogomolny

Draw ED'||BC, D' on AC. Let BD' meet CE in P. Since ΔBCP is equilateral,

∠CPD = 80°,

∠DPD' = 40°.

Since ∠DD'P = 40°, ΔDPD' is isosceles and

### Reference

- H. S. M. Coxeter, S. L. Greitzer,
*Geometry Revisited*, MAA, 1967, p. 159 - C. Knop,
__Nine Solutions to One Problem__,*Kvant*, 1993, no 6, Solution 1.

### 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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