 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  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 Comments: ( Back to comment list | You must be logged in to post comments.) Solution Comment 1 of 1
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`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 Please log in:

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