$4 (P M^{2} + R N^{2}) = 5 P R^{2}$

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Solution

In $△ P Q M$
${P M}^{2} = {P Q}^{2} + {Q M}^{2}$ --(1)
In $△ N Q R$
${N R}^{2} = {Q N}^{2} + {Q R}^{2}$ --(2)

Adding (1) and (2)

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

But $Q N = \frac{1}{2} Q P$ and $Q M = \frac{1}{2} Q R$

$∴ {Q N}^{2} = \frac{1}{4} {Q P}^{2}$ and ${Q M}^{2} = \frac{1}{4} {Q R}^{2}$

So, ${P M}^{2} + {N R}^{2} = {P Q}^{2} + \frac{1}{4} {Q R}^{2} + \frac{1}{4} {Q P}^{2} + {Q R}^{2}$

${P M}^{2} + {N R}^{2} = \frac{5}{4} ({P Q}^{2} + {Q R}^{2})$

$4 ({P M}^{2} + {N R}^{2}) = 5 ({P Q}^{2} + {Q R}^{2})$

$4 ({P M}^{2} + {N R}^{2}) = 5 (P R^{2})$

Hence proved.

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4 (P M^{2} + R N^{2}) = 5 Q R^{2}

4 P M^{2} = 4 P Q^{2} + Q R^{2}

P M^{2} + R N^{2} = 5 M N^{2}

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