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In a right triangle ABC, right angled at C, P and Q are points of the sides CA and CB respectively, which divide these sides in the ratio 2: 1. Prove that
9AQ2= 9AC2 +4BC2
9BP2= 9BC2 + 4AC2
9 (AQ2+BP2) = 13AB2
Ans: Since P divides AC in the ratio 2 : 1
CP = 2/3 AC QC = 2/3 BC
AQ2 = QC2 + AC2
AQ2 = 4/9 BC2 + AC2
9 AQ2 = 4 BC2 + 9AC2 ...................... (1)
Similarly we get 9 BP2 = 9BC2 + 4AC2 ....................(2)
Adding (1) and (2) we get 9(AQ2 + BP2) = 13AB2
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