In the figure, triangles ABC, DEF, and GHJ are congruent isosceles triangles. Angles B and C are 80 degrees. Points AEFC, EHJD, DFB, and GHF are collinear. Without drawing a circle, show how to mark the corners of a perfect hexagon on the diagram, using only a compass set to a constant length. Prove it's a perfect hexagon.
Source: Original.
This is my solution. Using the figure above. Angle BFC=EFD=80°. CBF is isosceles. Angle CBF is 20°. Angle ABF is 60°. With BK=BF, BFK is equilateral. Angle AFK is 40°. Similarly, FEH is isosceles and angle EFH is 20°. Angle KFH is 60°, and KFH is equilateral. Angle AKH is 60°. With KL=KH, KLH is equilateral. Angle HFD is 60°. With FM=FH, HFM is equilateral. Angle MHD is 40°. With HN=HJ, NHJ is isosceles. Angle NHJ is 20°, and MHN is equilateral. Angle NHG is 60°. With HO=HN, NHO is equilateral. We've shown five neighboring equilateral triangles, so angle LHO is 60° and LHO is equilateral. Six equilateral triangles make a perfect hexagon.