Question

In the given 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)

Answer 1

Construction: Through T, draw seg AB II side SR such that A-T-B, P- A-S and Q-B-R.

Proof:

seg PS || seg QR   ...(Opposite sides of rectangle)

∴ seg AS || seg BR   ...(P-A-S and Q-B-R)

also seg AB || seg SR   ...(Construction)

∴ `square` ASRB is a parallelogram   ...((By definition)

∠ASR = 90°   ...(Angle of rectangle PSRQ)

∴ `square` ASRB is a rectangle   ...(A parallelogram is a rectangle, if one of its angles is a right angle.)

∠SAB = ∠ABR = 90°   ...(Angle of a rectangle)

∴ seg TA ⊥ side PS and seg TB ⊥ side QR   ...(1)

AS = BR   ...(2) (Opposite sides of rectangle are equal)

Similarly, we can prove AP= BQ   ...(3)

In ΔTAS,

∠TAS = 90°   ...[From (1)]

∴ by Pythagoras theorem,

TS² = TA² + AS²   ...(4)

In ΔTBQ,

∠TBQ = 90°   ...[From (1)]

∴ by Pythagoras theorem,

TQ² = TB² + BQ²   ...(5)

Adding (4) and (5), we get,

TS² + TQ² = TA² + AS² + TB² + BQ²   ...(6)

In ΔTAP,

∠TAP = 90°   ...[From (1)]

∴ by Pythagoras theorem,

TP² = TA² + AP²   ...(7)

In ΔTBR,

∠TBR = 90°   ...[From (1)]

∴ by Pythagoras theorem,

TR² = TB² + BR²   ...(8)

Adding (7) and (8), we get

TP² + TR² = TA² + AP² + TB² + BR²

∴ TP² + TR² = TA² + BQ² + TB² + AS²   ...(9) [From (2) and (3)]

∴ from (6) and (g), we get,

TS² + TQ² = TP² + TR²

Answer 2

Construction: Draw diagonals PR and QS and let them intersect at 0. Draw seg TO.

Proof: `square` PQRS is a rectangle   ...(Given)

∴ PR=QS   ...(Diagonals of rectangle are congruent)

Multiplying both the sides by `1/2`, we get,

`1/2`PR = `1/2`QS   ...(1)

But, `1/2`PR = OP = OR   ...(2)

and `1/2`QS = OS = OQ   ...(3) [Diagonals of rectangle bisect each other]

∴ OP = OR = OS = OQ   ...(4) [From (1), (2) and (3)]

In ΔTSQ,

seg TO is the median   ...(By definition)

∴ by Apollonius theorem,

TS² + TQ² = 2TO² + 2OQ²   ...(5)

In ΔPTR,

seg TO is the median   ...(By definition)

∴ by Apollonius theorem,

TP² + TR² = 2TO² + 2OR²

∴ TP² + TR² = 2TO² + 2OQ²   ...(6) [From (4)]

∴ from (5) and (6), we get,

TS² + TQ² = TP² + TR²