# The 80-80-20 Triangle Problem, Solution #4

Let ABC be an isosceles triangle

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Copyright © 1996-2018 Alexander Bogomolny

This is Solution 3 from [Knop].

Draw a line through E parallel to BC and another through C parallel to AB. The two intersect in, say, H forming a parallelogram BCHE. Let P be the point on CE that completes the equilateral triangle BCP.

Then BP = BC = CD and BE = CH.

∠CHD = ∠BEC = 40°.

And, since ∠CHE = 80°, HD is the bisector of angle CHE. On the other hand, CD is the bisector of angle ECH. The point D is, therefore, the incenter of ΔECH, the intersection of its angle bisectors. So DE is the bisector of angle CEH and ∠CED = 30°.

### 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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