Proving a Convex Quadrilateral is a Square with Internal Point O as its Center

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SUMMARY

The discussion focuses on proving that a convex quadrilateral PQRS with an internal point O is a square if the condition 2A = OP² + OQ² + OR² + OS² holds true, where A is the area of the quadrilateral. The key to the proof lies in utilizing geometric principles such as the Pythagorean Theorem and the relationship between the segments of the quadrilateral. By analyzing the bisected segments and their squared lengths, one can establish the necessary conditions for PQRS to be a square with O as its center.

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Homework Statement



Let O be an internal point of a convex quaddrilateral PQRS whose area is A.

Prove that, if 2A = OP^2 + OQ^2 + OR^2 + OS^2, then PQRS is a square with O as its centre

Homework Equations





The Attempt at a Solution



I have no idea where to start, except that I know that I need to use 1/2 ab sin C
 
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Forget the 1/2 ab sin C for now. Draw a picture and think Pythagorean Theorum.
 
how does PQ, QR, RS and SP relate? Are they the same or different? If you bisect the segment PQ and square the result, how does that relate to the square of the segment OP? How about to the area of the square?
 

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