In the given figure, M is the midpoint of QR. ∠PRQ = 90°. Prove that, PQ²= 4PM²– 3PR²

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Solution

According to Pythagoras theorem,

In ∆PRM

${PR}^{2} + {RM}^{2} = {PM}^{2} \Rightarrow {RM}^{2} = {PM}^{2} - {PR}^{2} . . . (1)$

In ∆PRQ

${PR}^{2} + {RQ}^{2} = {PQ}^{2} \Rightarrow {PQ}^{2} = {PR}^{2} + {(RM + MQ)}^{2} \Rightarrow {PQ}^{2} = {PR}^{2} + {(RM + RM)}^{2} \Rightarrow {PQ}^{2} = {PR}^{2} + {(2 RM)}^{2} \Rightarrow {PQ}^{2} = {PR}^{2} + 4 {RM}^{2} \Rightarrow {PQ}^{2} = {PR}^{2} + 4 ({PM}^{2} - {PR}^{2}) (from (1)) \Rightarrow {PQ}^{2} = {PR}^{2} + 4 {PM}^{2} - 4 {PR}^{2} \Rightarrow {PQ}^{2} = 4 {PM}^{2} - 3 {PR}^{2}$

Hence, PQ²= 4PM²– 3PR².

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