Question

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

PQ² = 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 ∆PTQ, ∠PTQ = 90°   ...(Given)

∴ by Pythagoras theorem,

PQ² = PT² + TQ²

∴ PQ² = PT² + (QS − TS)²   ...(Q - T - S)

∴ PQ² = PT² + QS² − 2QS × TS + TS2   ...[(a − b)² = a² − 2ab + b²]

∴ PQ² = (PT² + TS²) − 2QS × TS + QS²

∴ PQ² = PS² − 2`(("QR")/2)` × TS + `(("QR")/2)^2`   ...[From (1) and (2)]

∴ PQ² = PS² − QR × ST + `(("QR")/2)^2`