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Question
In the given figure, seg PS is the median of ∆PQR and PT ⊥ QR. Prove that,
PR2 = PS2 + QR × ST + `("QR"/2)^2`
Solution
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`
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