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 

Answer 1

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.