#### Question

In the given figure, point T is in the interior of rectangle PQRS, Prove that, TS^{2 }+ TQ^{2 }= TP^{2 }+ TR^{2 }(As shown in the figure, draw seg AB || side SR and A-T-B)

#### Solution

According to Pythagoras theorem, in ∆PAT

In ∆ATS

In ∆QBT

In ∆BRT

Now,

\[{TS}^2 + {TQ}^2 = \left( {AS}^2 + {AT}^2 \right) + \left( {QB}^2 + {BT}^2 \right) \left( \text{From} \left( 2 \right) \text{and} \left( 3 \right) \right)\]

\[ \Rightarrow {TS}^2 + {TQ}^2 = \left( {BR}^2 + {AT}^2 \right) + \left( {PA}^2 + {BT}^2 \right) \left( \because AS = BR \text{ and } PA = QB \right)\]

\[ \Rightarrow {TS}^2 + {TQ}^2 = \left( {BR}^2 + {BT}^2 \right) + \left( {PA}^2 + {AT}^2 \right)\]

\[ \Rightarrow {TS}^2 + {TQ}^2 = \left( {TR}^2 \right) + \left( {PT}^2 \right) \left( \text{From} \left( 1 \right) \text{and} \left( 4 \right) \right)\]

\[ \Rightarrow {TS}^2 + {TQ}^2 = {TR}^2 + {PT}^2\]

Hence, TS^{2 }+ TQ^{2 }= TP^{2 }+ TR^{2}.