Prove which divide these sides in the ratio 2: 1, Mathematics

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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

1904_right triangle.png

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  + 9AC              ...................... (1)

Similarly we get 9 BP2 = 9BC2 + 4AC2 ....................(2)

Adding (1) and (2) we get 9(AQ2 + BP2) = 13AB2


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