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

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