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

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Solution

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

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

Adding (1) and (2),
${P M}^{2} + {R N}^{2} = {P Q}^{2} + {Q M}^{2} + {N Q}^{2} + {Q R}^{2} {P M}^{2} + {R N}^{2} = 4 {N Q}^{2} + 4 {Q M}^{2} + {M N}^{2}$ ...... $[2 P Q = N Q, 2 Q M = Q R]$
${P M}^{2} + {R N}^{2} = 5 {M N}^{2}$

Hence proved.

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