## The 80-80-20 Triangle Problem, A Derivative, Solution #6

ABC is an isosceles triangle with vertex angle

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

Copyright © 1996-2018 Alexander Bogomolny

This solution has been reported in [Leikin].

The proof is practically *by construction*.

Starting from BC, form a sequence of isosceles triangles, BCF, CFD, FDE, DEA. Is it possible? Let us check the angles.

For ΔBCF to be isosceles, suffice it to pick ∠BCF = 20°, for then

If point D on AC is such that DF = FC (and such point exists!), then, since

If E' on AB (not shown) is such that DF = DE' (and such a point exists), then ΔFDE' is isosceles,

It follows that ∠ADE' = 20° = ∠DAE'. Therefore, ΔADE' is isosceles and

Since ΔCFD is equilateral, CD = BC = DE, so that ΔCDE is also equilateral with

### Reference

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

Copyright © 1996-2018 Alexander Bogomolny

71531996