Question

in a right triangle ABC right angled at C,P and Q are the points on sides AB and AC respectively which divide these sides in the ratio 2:1. prove that
(i) 9AQ^2 = 9AC^2 + 4BC^2
(iI) 9BP^2 = 9BC^2 + 4AC^2
plsssss some1 solve this asaaap​

Answer 1

Step-by-step explanation:

Given : ΔABC is a right angle at C. P and Q are points on CA and CB respectively.

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

To prove :

1) 9AQ2 = 9AC2 + 4BC2

(2) 9BP2 = 9BC2 + 4AC2

(3) 9(AQ2 + BP2) = 13AB2

Proof : In a right angle ΔACQ,

AQ2 = AC2 + CQ2  [ CQ / QB = 2 /1 ,CQ / ( BC – CQ) = 2 / 1,3CQ = 2BC, CQ = 2BC / 3 ]

⇒ AQ2 = AC2 + (2BC / 3)2

⇒ AQ2 = AC2 + 4BC2 / 9

9 AQ2 = 9AC2 + 4BC2  . ---------(1)

Similarly in a right  angle ΔBCP we get

9BP2 = 9BC2 + 4AC2  ---------(2)

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  (∆ABC is right angle C then  AC 2  + BC 2  = AB 2 )