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Question
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(AQ2 + BP2) = 13AB2
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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(AQ2 + BP2) = 13(BC2 + AC2)
⇒ 9(AQ2 + BP2) = 13AB2.
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