Answer :

From figure,

In ∆PAT, ∠PAT = 90^{0}

TP^{2} = AT^{2} + PA^{2} …1

In ∆AST, ∠SAT = 90^{0}

TS^{2} = AT^{2} + SA^{2} …2

In ∆QBT, ∠QBT = 90^{0}

TQ^{2} = BT^{2} + QB^{2} …3

In ∆BTR, ∠RBT = 90^{0}

TR^{2} = BT^{2} + BR^{2} …4

TS^{2} + TQ^{2} = AT^{2} + SA^{2} + BT^{2} + QB^{2} [Adding 2 and 3]

⇒ TS^{2} + TQ^{2 =} AT^{2} + PA^{2} + BT^{2} + BR^{2} [SA = BR, QB = AP]

⇒ TS^{2} + TQ^{2} = TP^{2} + TR^{2} [From 1 and 4]

PROVED.

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