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Perpendicular Concurrency Concern (Posted on 2010-08-08) Difficulty: 3 of 5
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.

No Solution Yet Submitted by K Sengupta    
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Some Thoughts Limited progress | Comment 1 of 3
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
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