In a right angled triangle ABC, right angled at C, P and Q are the points on the sides CA and CB respectively which divide these sides in the ratio 2 : 1. Prove that

(i) 9AQ² = 9AC² + 4BC²

(ii) 9BP² = 9BC² + 4AC²

(iii) 9(AQ² + BP²) = 13AB²

(i) In right angle triangle ACQ,

By pythagoras theorem we get,

⇒ AQ² = AC² + CQ²

Multiplying both sides by 9 we get,

⇒ 9AQ² = 9AC² + 9CQ²

⇒ 9AQ² = 9AC² + (3CQ)² …….(i)

Given, BQ : CQ = 1 : 2

$\Rightarrow \dfrac{CQ}{BC} = \dfrac{CQ}{BQ + CQ} \\[1em] = \dfrac{2}{1 + 2} = \dfrac{2}{3} \\[1em] \Rightarrow \dfrac{CQ}{BC} = \dfrac{2}{3} \\[1em] \Rightarrow 3CQ = 2BC$

Substituting above value in (i) we get,

⇒ 9AQ² = 9AC² + (2BC)²

⇒ 9AQ² = 9AC² + 4BC².

Hence, proved that 9AQ² = 9AC² + 4BC².

(ii) In right angle triangle BPC,

By pythagoras theorem we get,

⇒ BP² = BC² + CP²

Multiplying both sides by 9 we get,

⇒ 9BP² = 9BC² + 9CP²

⇒ 9BP² = 9BC² + (3CP)² …….(ii)

Given, AP : PC = 1 : 2

$\Rightarrow \dfrac{CP}{AC} = \dfrac{CP}{AP + PC} \\[1em] = \dfrac{2}{1 + 2} = \dfrac{2}{3} \\[1em] \Rightarrow \dfrac{CP}{AC} = \dfrac{2}{3} \\[1em] \Rightarrow 3CP = 2AC$

Substituting above value in (ii) we get,

⇒ 9BP² = 9BC² + (2AC)²

⇒ 9BP² = 9BC² + 4AC².

Hence, proved that 9BP² = 9BC² + 4AC².

(iii) In right angle triangle ABC,

By pythagoras theorem we get,

⇒ AB² = AC² + BC² ……(iii)

Adding resultants from part (i) and (ii) we get,

9AQ² + 9BP² = 9AC² + 4BC² + 9BC² + 4AC²

9(AQ² + BP²) = 13AC² + 13BC²

9(AQ² + BP²) = 13(AC² + BC²)

9(AQ² + BP²) = 13AB² [From (iii)].

Hence, proved that 9(AQ² + BP²) = 13AB².