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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
concurrent.

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)

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education Comment 1 of 1
Flooble has a forum about shapes with theitlr gravity locks. They are giving details of their quadrilateral ABCD is theline segment MF. Here you can check this bestdissertation.com and learn more new tricks about college work. You can also get the maltitudes and Z the intersection of the diagonals plans. Join it for more.
  Posted by Supon1984 on 2021-03-05 22:48:14
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