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Four Equal Triangles (Posted on 2013-06-16) Difficulty: 3 of 5

Here's a problem I found in the Problems section of a
math journal. The way I read it, it is not true.

Let ABCD be a convex quadrilateral. Prove that

there exists a point P inside ABCD such that
[PAB]=[PBC]=[PCD]=[PDA],

    if and only if

the diagonals bisect each other.

Here [XYZ] denotes the area of triangle XYZ.

See The Solution Submitted by Bractals    
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Comments: ( Back to comment list | You must be logged in to post comments.)
re: Solution | Comment 2 of 3 |
(In reply to Solution by Jer)

I arrived at the same conclusion.


What about restating the problem by replacing
     "the diagonals bisect each other"
                  with
     "at least one diagonal bisects the other"?

  Posted by Bractals on 2013-06-17 13:56:07
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