ABCDEF is a convex, but not necessarily regular, hexagon with AB = BC; CD = DE; EF = FA and < ABC + < CDE + < EFA = 300o.
Prove that the perpendiculars from A, C and E respectively to FB, BD and DF are concurrent.
Has anyone else had a look at this problem using Cabri (or another geometry package)? It seems to me that the perpendiculars are concurrent whatever the sum of the three angles; but finding a concise proof is proving difficult. Perhaps the constraint is there to make it easier. Any thoughts?
Posted by Harry
on 2010-08-26 22:38:24