# The 80-80-20 Triangle Problem, Solution #7

Let ABC be an isosceles triangle

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Copyright © 1996-2018 Alexander Bogomolny

This is Solution 6 from [Knop] and is due to Maria Gelband, a high school student at the time.

Let M be on the bisector of angle ∠BCD with the property that

Since ΔBCD is isosceles (BC = CD), CM is the perpendicular bisector of BD so that,

Since BE is the bisector of ∠DBM it is also the perpendicular bisector of DM.

In ΔCDM, E is the intersection of the perpendicular bisector of DM and the bisector of the angle at C. Thus E belongs to the circumcenter of ΔCDM. In the circumcircle, two inscribed angles, CED and CMD are equal, but the latter is equal half of ∠BMD making

### 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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