Question

In the adjoining figure, point T is in the interior of rectangle PQRS. Prove that, TS² + TQ² = TP² + TR². 

(As shown in the figure, draw seg AB || side SR and A – T – B)

Given: PQRS is a rectangle. 

Point T is in the interior of PQRS. 

To prove: TS² + TQ² = TP² + TR²

Construction: Draw seg AB || side SR such that A – T – B.

Answer 1

Proof:

PQRS is a rectangle. [Given]

∴ PS = QR (i) [Opposite sides of a rectangle]

In ASRB,

∠S = ∠R = 90° (ii) [Angles of rectangle PQRS] side AB || side SR [Construction]

Also ∠A = ∠S = 90° [Interior angle theorem, from (ii)]

∠B = ∠R = 90° 

∴ ∠A = ∠B = ∠S = ∠R = 90° (iii)

∴ ASRB is a rectangle.

∴ AS = BR (iv) [Opposite sides of a rectangle

In ∆PTS, ∠PST is an acute angle and seg AT ⊥ side PS [From (iii)]

∴ TP² = PS² + TS²  – 2 PS. AS (v) [Application of Pythagoras theorem]

In ∆TQR., ∠TRQ is an acute angle and seg BT ⊥ side QR [From (iii)]

∴ TQ² = RQ² + TR² – 2 RQ. BR (vi) [Application of pythagoras theorem]