ABCDEF is a convex hexagon with all its sides equal. Also A + C + E = B + D + F. Show that A = D, B = E and C = F.
Solution
Construct a copy A'B'C'D'E'F' of ABCDEF with F'E' placed on BC as shown. Construct another copy A"...F" with A"B" placed on CD as shown. Since ∠A + ∠C + ∠E = 360o, D' and B" coincide. Take the lengths BF, FD, DB to be x, y, z as shown. Then the we have BD' = y and DD' = x as shown. So FBD and D'DB are congruent and ∠DBD' = ∠FDB. But ∠D'BB' = ∠BFD, so ∠FBD + ∠DBD' + ∠D'BB' = 180o, and FBB' is a straight line. But triangles AFB and ABB' are congruent, so AF is parallel to A'B. Hence FE is parallel to BC. Similarly, we can prove that the other pairs of opposite sides are parallel.
If we rotate AB through ∠A we get AF. If we rotate that through ∠F we get EF. If we rotate that through ∠E we get ED, which is parallel to AB. Hence ∠A + ∠F + ∠E = 360o. But ∠A + ∠C + ∠E = 360o, so ∠C = ∠F. Similarly for the other pairs of opposite angles.
© John Scholes
jscholes@kalva.demon.co.uk
29 Oct 2003
Last corrected/updated 29 Oct 03