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| Inscribe Quadrilateral, Get Diophantus (Posted on 2007-04-09) |
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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
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Submitted by K Sengupta
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Rating: 5.0000 (1 votes)
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Solution:
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(Hide)
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Let O correspond to the centre of the circle.
Then:
Angle POQ + Angle COD
= 2(Angle PRQ + Angle RQS)
= 2*90o
= 180o
Then:
PQ2 + RS2
= 2(T2 - T2*cos t)+ 2(T2 - T2*cos (pi -t))
= 4T2
In a similar manner, it can be shown that:
QR2 + PS2
= 4T2
Consequently:
HP2 + HQ2 + HR2 + HS2
= 1/2*(PQ2 + QR2 + RS 2+ SP2
= 1/2* 8S2
= 4T2
*******************************
Also refer to the solution posted by Bractals in this location.
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Subject |
Author |
Date |
 | Solution | Bractals | 2007-04-09 20:42:12 |
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