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Question
The perpendicular PS on the base QR of a ∆PQR intersects QR at S, such that QS = 3 SR. Prove that 2PQ2 = 2PR2 + QR2
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Solution
Given QS = 3SR
QR = QS + SR
= 3SR + SR = 4SR
SR = `1/4` QR ...(1)
QS = 3SR
SR = `"QS"/3` ...(2)
From (1) and (2) we get
`1/4 "QR" = "QS"/3`
∴ QS = `3/4` QR ...(3)
In the right ∆PQS,
PQ2 = PS2 + QS2 ...(4)
Similarly in ∆PSR
PR2 = PS2 + SR2 ...(5)
Subtract (4) and (5)
PQ2 – PR2 = PS2 + QS2 – PS2 – SR2
= QS2 – SR2
PQ2 – PR2 = `[3/4 "QR"]^2 - ["QR"/4]^2`
From (3) and (1)
= `(9"QR"^2)/16 - "QR"^2/16`
= `(8"QR"^2)/16`
PQ2 – PR2 = `1/2 "QR"^2`
2PQ2 – 2PR2 = QR2
2PQ2 = 2PR2 + QR2
Hence the proved.
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