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 

Answer 1

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)