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 Inscribe Quadrilateral, Get Diophantus (Posted on 2007-04-09)
PQRS is a cyclic quadrilateral with PR perpendicular to QS while PR meets QS at H. T is the radius of the circumscribed circle.

Prove that:

HP2 + HQ2 + HR2 + HS2 = 4T2

 See The Solution Submitted by K Sengupta No Rating

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 Solution Comment 1 of 1

Let O be the center of the circumscribed circle.
HP^2 + HQ^2 + HR^2 + HS^2
= PQ^2 + RS^2
= OP^2 + OQ^2 - 2(OP)(OQ)cos(POQ) +
OR^2 + OS^2 - 2(OR)(OS)cos(ROS)
= 4T^2 - 2T^2[cos(POQ) + cos(ROS)]
= 4T^2 - 2T^2[cos(2PRQ) + cos(2RQS)]
= 4T^2 - 2T^2[cos(2HRQ) + cos(2RQH)]
= 4T^2 - 2T^2[cos(2HRQ) + cos(2[90-HRQ])]
= 4T^2 - 2T^2[cos(2HRQ) + cos(180 - 2HRQ)]
= 4T^2 - 2T^2[cos(2HRQ) - cos(2HRQ)]
= 4T^2

 Posted by Bractals on 2007-04-09 20:42:12

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