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

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)

Applying Pythagoras theorem in right triangle BCP, we have
BP² = BC² + CP²

⇒ BP² = `"BC"^2 + (4)/(9)"AC"^2`   ...(Using (i))

⇒ 9BP² = 9BC² + 4AC².