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 + 9AC² ...................... (1)

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

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

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