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The perpendicular PS on the base QR of a ∆PQR intersects QR at S, such that QS = 3 SR. Prove that 2PQ^{2} = 2PR^{2} + QR^{2}

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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,

PQ^{2} = PS^{2} + QS^{2} ...(4)

Similarly in ∆PSR

PR^{2} = PS^{2} + SR^{2} ...(5)

Subtract (4) and (5)

PQ^{2} – PR^{2} = PS^{2} + QS^{2} – PS^{2} – SR^{2}

= QS^{2} – SR^{2}

PQ^{2} – PR^{2} = `[3/4 "QR"]^2 - ["QR"/4]^2`

From (3) and (1)

= `(9"QR"^2)/16 - "QR"^2/16`

= `(8"QR"^2)/16`

PQ^{2} – PR^{2} = `1/2 "QR"^2`

2PQ^{2} – 2PR^{2} = QR^{2}

2PQ^{2} = 2PR^{2} + QR^{2}

Hence the proved.

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