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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