All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Shapes > Geometry
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    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
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

Please log in:
Remember me:
Sign up! | Forgot password

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (0)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2021 by Animus Pactum Consulting. All rights reserved. Privacy Information