Question

In the adjoining figure, $△PQR$ is right angle at $Q$ and points $S$ and $T$ trisect side $QR$. Prove that
$8{PT}^2=3{PR}^2+5{PS}^2$

Answer 1

Let $QS=x$, then,

$QT=2x$

$QR=3x$

In $\Delta PQR$,

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

$P{R}^2=P{Q}^2+{\left(3x\right)}^2$

$P{R}^2=P{Q}^2+9x^2$ (1)

In $\Delta PQT$,

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

$P{T}^2=P{Q}^2+{\left(2x\right)}^2$

$P{T}^2=P{Q}^2+4x^2$ (2)

In $\Delta PQS$,

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

$P{S}^2=P{Q}^2+x^2$ (3)

Now, using equation (1) and (3),

$3P{R}^2+5P{S}^2=3\left(P{Q}^2+9x^2\right)+5\left(P{Q}^2+x^2\right)$

$=3P{Q}^2+27x^2+5P{Q}^2+5x^2$

$=8P{Q}^2+32x^2$

$=8\left(P{Q}^2+4x^2\right)$

From equation (2),

$3P{R}^2+5P{S}^2=8P{T}^2$

Hence proved.