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

(1)

(2)

According to the question,

QS = SR = , ∠T = 90⁰

Now in ∆PTR, ∠PTR = 90⁰

PT² + TR² = PR²

⇒ PR² = PT² + (ST + )²

⇒ PR² = PT² + (ST + )²

⇒ PR² = PT² + ST² + 2.ST. – – – – 1

Similarly in ∆PTS

PS² = PT² + ST² – – – – 2

1 and 2 implies,

PR² = PS² – ST² + ST² + 2.ST.

⇒ PR² = PS² + ST.QR +

PROVED.

Now in ∆PTQ, ∠PTQ = 90⁰

PT² + TQ² = PQ²

⇒ PR² = PT² + (ST + )²

⇒ PR² = PT² + ( – ST)²

⇒ PR² = PT² + ST² – 2.ST. …1

Similarly in ∆PTS

PS² = PT² + ST² …2

1 and 2 implies,

PR² = PS² – ST² + ST² – 2.ST.

⇒ PR² = PS² – ST.QR +

PROVED.