If $S$ is a point on the side $P Q$ of $△ P Q R$ such that $P S = Q S = R S$ , then

P R . Q R = {R S}^{2}

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{Q S}^{2} + {R S}^{2} = {Q R}^{2}

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{P R}^{2} + {Q R}^{2} = {P Q}^{2}

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{P S}^{2} + {R S}^{2} = {P R}^{2}

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Solution

The correct option is C ${P R}^{2} + {Q R}^{2} = {P Q}^{2}$
Given: $△ P Q R$ , $P S = Q S = R S$
Now, In $△ P R S$ ,
$P S = R S$
thus, $∠ S P R = ∠ S R P = x$ ...(Isosceles triangle property)
Now, In $△ S Q R$ ,
$S Q = Q R$
$∠ S Q R = ∠ S R Q = y$ ...(Isosceles triangle property)
Sum of angles of PQR = 180
$∠ P Q R + ∠ P R Q + ∠ Q P R = 180$
$∠ P Q R + ∠ Q P R + ∠ Q R S + ∠ P R S = 180$
$x + y + x + y = 180$
$x + y = 90$
$∠ P R S + ∠ Q R S = 90$
$∠ P R Q = 90^{\circ}$
Thus, $△ P Q R$ , is a right angled triangle at $∠ R$
Thus, $P R^{2} + Q R^{2} = P Q^{2}$ ...(Pythagoras Theorem)

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Δ

(A) (P R) (Q R) = R S^{2}

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(C) P R^{2} + Q R^{2} = P Q^{2}

(D) P S^{2} + R S^{2} = P R^{2}

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90^{o}

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