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In a triangle ABC right angled at C, P and Q are points of sides CA and CB respectively, which divide these sides the ratio 2 : 1.

Prove that : 9(AQ^{2} + BP^{2}) = 13AB^{2}

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#### Solution

P divides AC in the ratio 2 : 1

So C.P. = `(2)/(3) "AC"` .......(i)

Q divides BC in the ratio 2 : 1

QC = `(2)/(3)"BC"` ......(ii)

Adding (iii) and (iv), we get

9(AQ^{2 }+ BP^{2}) = 13(BC^{2} + AC^{2})

⇒ 9(AQ^{2} + BP^{2}) = 13AB^{2}.

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