Q5 of 45 Page 43

In figure 2.30, point T is in theinterior of rectangle PQRS, Prove that, TS2 + TQ2 = TP2 + TR2(As shown in the figure, drawseg AB || side SR and A – T – B)

From figure,


In ∆PAT, ∠PAT = 900


TP2 = AT2 + PA2 …1


In ∆AST, ∠SAT = 900


TS2 = AT2 + SA2 …2


In ∆QBT, ∠QBT = 900


TQ2 = BT2 + QB2 …3


In ∆BTR, ∠RBT = 900


TR2 = BT2 + BR2 …4


TS2 + TQ2 = AT2 + SA2 + BT2 + QB2 [Adding 2 and 3]


⇒ TS2 + TQ2 = AT2 + PA2 + BT2 + BR2 [SA = BR, QB = AP]


⇒ TS2 + TQ2 = TP2 + TR2 [From 1 and 4]


PROVED.


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