$M$ is the midpoint of side $Q R$ of $△ P Q R . P S ⊥ Q R$ Prove that: $P Q^{2} + P R^{2} = 2 P M^{2} + 2 Q M^{2}$

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Solution

M is the midpoint, $\Rightarrow Q M = M R$

$△ Q P S$

$P Q^{2} = P S^{2} + Q S^{2}$ ...(1)

$△ R S P$

$P R^{2} = P S^{2} + S R^{2}$ ...(2)

Adding (1) and (2), we get,

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

$= 2 P M^{2} + Q S^{2} + S R^{2}$

$∴ P Q^{2} + P R^{2} = 2 P M^{2} + 2 Q M^{2}$

Hence proved.

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