# The 80-80-20 Triangle Problem, Solution #8

Let ABC be an isosceles triangle

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

Copyright © 1996-2018 Alexander Bogomolny

This is Solution 7 from [Knop] and is due to Alexander Kornienko, a high school student at the time.

Reflect B in AC and C in AB to obtain B' and, respectively, C'.

Now, ∠AB'D = ∠ABD = 30°. Further,

Since ∠AC'E = ∠ACE = ∠CAE = ∠C'AE, we see that E is equidistant from A, C, and C'. E then lies on the perpendicular bisector of AC', i.e. on B'D. But then

∠CED = 180° - 40° - 40° - 70° = 30°.

(This solution is neatly embedded into a configuration of an 18-gon.)

### 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

64995017 |