in a right triangle ABC , right angled at C; P and Q are points of the sides AC and BC respectively , which divides these sides in the ratio 2:1. prove that :

(a)9(AO)2=9(AC)2 + 4(BC)2

(b)9(BP)2 =9(BC)2 + 4(AC)2

(c)9[(AO)2 + (BP)2] = 13(AB)2

Given : ∆ABC is a right triangle. ∠ACB = 90⁰. P and Q are points on CA and CB respectively.

CP: PA = 2:1 and CQ: QB = 2:1

To Prove :

(a) 9AQ² = 9AC² + 4BC²

(b) 9BP² = 9BC² + 4AC²

(c) 9(AQ² + BP²) = 13AB²

Proof : 

In right ∆ACQ,

AQ² = AC² + CQ²

Similarly,

In right ∆BCP,

9BP² = 9BC² + 4AC² .........................(2)

Now, 

on adding (1) and (2), we get,

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

⇒ 9(AQ ² + BP ² ) = 13AC ² + 13BC ²

⇒ 9(AQ ² + BP ² ) = 13(AC ² + BC ² ) = 13AB ²   (In right ∆ABC, AC ² + BC ² = AB ² )