Question

In the given figure, seg PS is the median of ∆PQR and PT ⊥ QR. Prove that,

PR² = PS² + QR × ST + `("QR"/2)^2`

Answer 1

Seg PS is the median of ∆PQR   ...(Given)

∴ QS = SR = `1/2`QR   ...(1) [S is the midpoint of side QR]

In ∆PTS, ∠PTS = 90°   ...(Given)

∴ by Pythagoras theorem,

PS² = PT² + TS²   ...(2)

In ∆PTR, ∠PTR = 90°   ...(Given)

∴ by Pythagoras theorem,

PR² = PT² + TR²

∴ PR² = PT² + (TS + SR)²   ...(T-S-R)

∴ PR² = PT² + TS² + 2ST2.SR + SR²   ...[(a + b)² = a² + 2ab + b²]

∴ PR² = (PT² + TS²) + 2ST.SR + SR²

∴ PR² = PS² + 2ST.`(("QR")/2) + (("QR")/2)^2`   ...[From (1) and (2)]

∴ PR² = PS² + QR × ST + `(("QR")/2)^2`