Question

In $\Delta$ PQR, $\angle$ PQR $=90$. X and Y are the points of PQ and QR such that PX$:$XQ$=1:2$ and QY$:$YR$=2:1$ prove that $9(P{Y}^2+X{R}^2)=13P{R}^2$.

Answer 1

We are given that,

$In△PQR\angle PQR={90}^o$

Also, $In△XQY,\angle XQY={90}^o$

Similarly in $△PQY$ & $△XQR,\angle PQY=\angle XQR={90}^o$

Let $PQ=3a$ So, $PX=a$ & $XQ=2a$ & $QR=3b$ & $QY=2b$ & $YR=b$

Where $a$ & $b$ are any positive real no. $a,b\in R^{+}$

here from the Pythagoras theorem we get.

$P{R}^2=P{Q}^2+Q{R}^2\to \left(1\right)$

$X{Y}^2=X{Q}^2+Q{Y}^2\to \left(2\right)$

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

& $X{R}^2=X{Q}^2+Q{R}^2\to \left(4\right)$

So, $P{R}^2=g(a^2+b^2)$

From condition

here $PR,XY,PY$ & $XR$ are diagonal

So, $LHS=9(P{Y}^2+X{R}^2)$

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

From $\left(3\right)$ and $\left(4\right)$

$=9(9a^2+4b^2+4a^2+9b^2)$

$=9(13a^2+13b^2)$

$=\left(13\right)\left(9\right)(a^2+b^2)$

$=13.P{R}^2$

$=RHS.LHS=RHS$