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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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