Constructing a Regular Hexagon from Triangles

[BMAD]

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.

[Rainman]

I have a solution if I can use a ruler as well as the compass, and the compass is set to the length of BC. Also my solution uses the parallel postulate.

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.

[bonanova]

Cool. I didn't find the initial 120^o angle.

[plainglazed]

set compass 1/3 of length HF and strike a point M on FD from point F using compass three times like a divider. This gives equilateral triangle FHM. set compass on F, H, and M and strike the two corresponding segments of equilateral triangle FHM gives the six points of a hexagon. This method still requires a straight edge to connect FH and FM. Not sure from the OP if that is allowed. Or if striking an arc would be allowed (vs drawing a circle). If no ruler is allowed, I cannot visualize where a hexagon might be much less locate it.

[BMAD]

yeah, i took it for granted that we could draw a straight line...