Question

In a right angled triangle ABC, right angled at C, P and Q are the points on the side CA and C respectively which divide these sides in the ratio 2 : 1. Prove that
(i) $9A{Q}^2=9A{C}^2+4B{C}^2$
(ii) $9B{P}^2=9B{C}^2+4A{C}^2$
(iii) $9\left(A{Q}^2+B{P}^2\right)=13A{B}^2$

Answer 1

In $ACQ$

$\left(1\right)A{Q}^2=A{C}^2+Q{C}^2\frac{CQ}{QB}=\frac{1}{2}\&\frac{CP}{CA}=\frac{2}{1}$

$\frac{CQ}{QB}=\frac{2}{1}$

$\Rightarrow \frac{CQ}{BC-CQ}=2$

$\Rightarrow 3CQ=2BC$

$\Rightarrow CQ=\frac{2}{3}BC$

By $\left(1\right)$

$A{Q}^2=A{C}^2+{\left(\frac{2}{3}BC\right)}^2$

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

In $BCP\frac{CP}{CA-CP}=2$

$P{B}^2=B{C}^2+P{C}^2--CP=\frac{2}{3}CA$

$P{B}^2=B{C}^2+{\left(\frac{2}{3}CA\right)}^2$

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

$\left(3\right)$ Adding $\left(A\right)\&\left(B\right)$

$9A{Q}^2=9B{C}^2=13\left(B{C}^2+A{C}^2\right)$

$\Rightarrow 9\left(A{Q}^2+B{P}^2\right)=13A{B}^2$