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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: 9AQ2 = 9AC2 + 4BC2
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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)
In ΔACQ
Using Pythagoras Theorem we have,
AQ2 + AC2 + CQ2
⇒ AQ2 = `"AC"^2 + (4)/(9)"BC"^2` ...(using (ii))
⇒ 9AQ2 = 9AC2 + 4BC2. ......(iii)
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