In the figure 2.28 seg PS is the median of PQR and PT⊥QR. Prove that,

(1) PR2 = PS2 + QR × ST +( QR/ 2 )2
(ii) PQ2 = PS2 – QR ×ST +( QR /2)2


(1) QS = ½ QR ……(i) [S is the midpoint of QR]
SR = ½ QR ……(ii)
QS = SR [From (i) and (ii)]
PT ⊥QR [Given]
PSR is an obtuse angle. [From figure]
PR2 = SR2+PS2+2SR×ST …..(iii) [Application of Pythagoras theorem]
Substitute SR = ½ QR in (iii)
PR2 = PS2 + QR×ST +(QR/ 2 )2
Hence proved.

(ii) PT⊥QS [Given]

PSQ is an acute angle [From figure]
PQ2 = QS2+PS2-2QS×ST [Application of Pythagoras theorem]
PR2 = [(1/2)QR]2+PS2-2(1/2 )QR×ST
PR2 = (QR/2)2+PS2-QR×ST
PR2 = PS2-QR×ST+( QR /2)2
Hence proved.

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