Best Answer Rainman, 16 November 2013 - 08:13 PM

Since E-H-J and G-H-F, angle EHF = angle GHJ = 80^{o}. So the triangle EFH also has base angles of 80^{o}, which makes it isosceles. Similarly, the triangle CBF is also isosceles, with base angles of 80^{o}. The top angles must be 20^{o} accordingly. Now, angle HFB = angle HFE + angle EFB = 20^{o} + (180^{o} - angle EFD) = 20^{o} + 100^{o} = 120^{o}, which is the internal angle of a perfect hexagon.

Centering the compass at B, find the point M on the segment BA so that |BM| = |BC| (which is the set length of the compass). Consider the triangle MBF. Clearly |BM| = |BC| = |BF|, so it is isosceles, with top angle MBF = angle MBC - angle FBC = 80^{o }- 20^{o} = 60^{o}. An isosceles triangle with a top angle of 60^{o} is equilateral. It follows that M is the midpoint of the perfect hexagon with B,F,H as three of the corners. With a ruler, extend the lines BM-, FM-, and HM-. Using the compass centered at M, find the other three corners on those lines.