Question

In a right 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 [4 MARKS]

(i) $9A{Q}^2=9{AC}^2+4{BC}^2$

(ii) $9{BP}^2=9{BC}^2+4{AC}^2$

(iii) $9({AQ}^2+{BP}^2)=13{AB}^2$

Answer 1

Concept: 1 Mark
Each proof : 1 Mark each




It is given that P divides CA in the ratio 2 : 1. Therefore,

$CP=\frac{2}{3}AC\dots \dots \left(1\right)$

Also, Q divides CB in the ratio 2 : 1

$\therefore QC=\frac{2}{3}BC\dots \dots \left(2\right)$


(i) Applying Pythagoras theorem in right-angled triangle ACQ, we have

$A{Q}^2=Q{C}^2+A{C}^2$

$\Rightarrow A{Q}^2=\frac{4}{9}B{C}^2+A{C}^2$ [Using (2)]

$\Rightarrow 9A{Q}^2=4B{C}^2+9A{C}^2\dots \dots \left(3\right)$

(ii) Applying Pythagoras theorem in right triangle BCP, we have

$B{P}^2=B{C}^2+C{P}^2$

$\Rightarrow B{P}^2=B{C}^2+\frac{4}{9}A{C}^2$ [Using (1)]

$\Rightarrow 9B{P}^2=9B{C}^2+4A{C}^2\dots \dots \left(4\right)$

(iii) Adding (3) and (4), we get

$9(A{Q}^2+B{P}^2)=13(B{C}^2+A{C}^2)$

$\Rightarrow 9(A{Q}^2+B{P}^2)=13A{B}^2$ $[\because B{C}^2=A{C}^2+A{B}^2]$