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

Home > Shapes > Geometry
Orthodiagonal Quadrilateral (Posted on 2013-03-24) Difficulty: 3 of 5

A maltitude (midpoint-altitude) of quadrilateral ABCD is the
line segment MF (where M is the midpoint of side AB and F is
the foot of the perpendicular from M to the line CD). The 
other three maltitudes are defined similarly.

Prove that the maltitudes of a cyclic quadrilateral are

Let ABCD be a cyclic quadrilateral that is not a rectangle.
Let T be the point of concurrency of the maltitudes and 
Z the intersection of the diagonals. Prove the follwing:

            T = Z  ⇔  AC ⊥ BD.

See The Solution Submitted by Bractals    
Rating: 3.0000 (1 votes)

Comments: ( You must be logged in to post comments.)
  Subject Author Date
There are no comments yet.
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 (4)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information