ABCDEF is a convex, but not necessarily regular, hexagon with AB = BC; CD = DE; EF = FA and < ABC + < CDE + < EFA = 300^o.

Prove that the perpendiculars from A, C and E respectively to FB, BD and DF are concurrent.

(In reply to Solution by Harry)

I like your proof and I agree that the 300 degree restriction is not necessary.

Posted by Bractals on 2010-08-28 14:19:36