Question

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

i. PR² = PS² + QR × ST + ( QR/2)² 

ii. PQ² = PS² – QR × ST + (QR/2)²

Answer 1

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

∴ In ∆PSR, ∠PSR is an obtuse angle [Given] 

and PT ⊥ SR [Given, Q-S-R] 

∴ PR² = SR² +PS² + 2 SR × ST (ii) [Application of Pythagoras theorem] 

∴ PR² = ( 1/2QR)² + PS² + 2 ( 1/2 QR) × ST [From (i) and (ii)] 

∴ PR² = (QR/2 )² + PS² + QR × ST 

∴ PR² = PS² + QR × ST + (QR/2)²

ii. In.∆PQS, ∠PSQ is an acute angle and [Given]

PT ⊥QS [Given, Q-S-R] 

∴ PQ² = QS² + PS² – 2 QS × ST (iii) [Application of Pythagoras theorem] 

∴ PR² = ( 1/2 QR)² + PS² – 2 ( 1/2QR) × ST [From (i) and (iii)] 

∴ PR² = (QR/2)² + PS² – QR × ST 

∴ PR² = PS² – QR × ST + (QR/2)²