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 : 9(AQ² + BP²) = 13AB² 

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)

Adding (iii) and (iv), we get
9(AQ² + BP²) = 13(BC² + AC²)
⇒ 9(AQ² + BP²) = 13AB².