Question

In the figure, $\Delta$PQR is an equilateral triangle. Point S is on seg OR such that $QS=\frac{1}{3}$QR. Prove that: $9P{S}^2=7P{Q}^2$.

Answer 1

$PT$ be the perpendicular from $P$ on $QR$. Hence,$QT=\frac{1}{2}QR$

$ST=QR(\frac{1}{2}-\frac{1}{3})=\frac{1}{6}QR$

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

$also,PT=QT\tan {60}^{\circ }$

$PT=\sqrt{3}QT$

$P{S}^2=3Q{T}^2+\frac{Q{R}^2}{36}$

$P{S}^2=\frac{3Q{R}^2}{4}+\frac{Q{R}^2}{36}$

$P{S}^2=\frac{28Q{R}^2}{36}=\frac{7Q{R}^2}{9}$

$QR=PQ$

$9P{S}^2=7P{Q}^2$