Question

In figure, $PQR$ is a triangle, right angled at $Q$. $X$ and $Y$ are the points on $PQ$ and $QR$ such that $PX:XQ=1:2$ and $QY:YR=2:1$. Prove that $9({PY}^2+{XR}^2)={13PR}^2$.

Answer 1

Given that $X$ divides $PQ$ in $2:1$

So, $QX=\frac{2}{3}PQ..\left(1\right)$

Similarly $Y$ divides $QR$ in $2:1$

So, $QY=\frac{2}{3}QR..\left(1\right)$

In $△PYQ$

$P{Y}^2=P{Q}^2+Q{Y}^2$

$\Rightarrow P{Y}^2=P{Q}^2+{\left(\frac{2}{3}QR\right)}^2$

$\Rightarrow 9P{Y}^2=9P{Q}^2+4Q{R}^2........\left(3\right)$

In $△XQR$

$X{R}^2=Q{X}^2+Q{R}^2$

$\Rightarrow X{R}^2={\left(\frac{2}{3}PQ\right)}^2+Q{R}^2$

$\Rightarrow 9X{R}^2=4P{Q}^2+9Q{R}^2........\left(4\right)$

On adding (3) and (4), we get

$9(P{Y}^2+X{R}^2)=13(P{Q}^2+Q{R}^2)$

$9(P{Y}^2+X{R}^2)=13P{R}^2$

Hence proved.