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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: 9BP2 = 9BC2 + 4AC2
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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)
Applying Pythagoras theorem in right triangle BCP, we have
BP2 = BC2 + CP2
⇒ BP2 = `"BC"^2 + (4)/(9)"AC"^2` ...(Using (i))
⇒ 9BP2 = 9BC2 + 4AC2.
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