PQR is a right angle triangle, right angled at Q if QS = SR. Prove that $P R^{2} = 4 P S^{2} - 3 P Q^{2}$

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Solution

$I n Δ P Q R, P R^{2} = P Q^{2} + Q R^{2} Q R = Q S + S R Q S = Q R (G i v e n)$
Therefore,
$P R^{2} + P Q^{2} + (Q S + Q R)^{2} P R^{2} = P Q^{2} + (2 Q S)^{2} P R^{2} = P Q^{2} + 4 Q S^{2} . - - - - - - - (i) I n Δ P Q S, P S^{2} + P Q^{2} + Q S^{2} Q S^{2} = P S^{2} - P Q^{2}$
Putting the values of QS^2 in eq. (i), we get
$P R^{2} = P Q^{2} + 4 (P S^{2} - P Q^{2}) P R^{2} = P Q^{2} + 4 P S^{2} - 4 P Q^{2} P R^{2} = 4 P S^{2} - 3 P Q^{2}$

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In a triangle PQR right angled at Q if QS = SR then prove that PR² = 4PS²- PQ²

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