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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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re: Solution Comment 3 of 3 |
(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

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