In Fig. Point T is in the interior of rectangle PQRS, Prove that $T S^{2} + T Q^{2} = T P^{2} + T R^{2}$ . ( As shown in the figure, draw seg AB||SR and A-T-B

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Solution

Draw $A B ⊥$ to $P Q$ and $S R$ passes through $T$
In $Δ P A T$ & $Δ Q A T$
$P T^{2} = P A^{2} + A T^{2}$
$\Rightarrow P T^{2} = P A^{2} + A T^{2} . . . . . . . . . . . (1)$
$Q T^{2} = Q A^{2} + A T^{2} . . . . . . . . . . . . (1)$
$\Rightarrow Q T^{2} - Q A^{2} = A T^{2} . . . . . . . . . . . . . . . (2)$
from $(1)$ & $(2)$
$P T^{2} - P A^{2} = Q T^{2} - Q A^{2} . . . . . . . . . . . . . (3)$
Now in $Δ S B T$ and $Δ R B^{'} T$
$S T^{2} = S B^{2} + T B^{2}$
$S T^{2} - S B^{2} = T B^{2} . . . . . . . . . . . . . . . . . . . . (4)$
$T R^{2} = T B^{2} = B R^{2}$
$\Rightarrow T R^{2} - B^{1} R^{2} = T B^{2} . . . . . . . . . . . . . . . . . . (5)$
From $(4)$ & $(5)$
$S T^{2} - S B^{2} = T R^{2} - B R^{2} . . . . . . . . . . . . . . . (6)$
From $(3)$ & $(6)$ , $∵ P A = S B$ and $A Q = B R$
$P T^{2} + R T^{2} = Q T^{2} + S T^{2}$

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