Question

In this fig. $M$ is the midpoint of $QR$, $\angle PRQ={90}^{\circ }$, then prove that, $P{Q}^2=4P{M}^2-3P{R}^2$

Answer 1

To prove : ${PQ}^2={4PM}^2-3{PR}^2$

Proof :-

In $\Delta PRM$

${PM}^2={PR}^2+{RM}^2$ (Pythagoras theorem) $⟶\left(1\right)$

Also,

In $\Delta PQR$

${PQ}^2={PR}^2+{RQ}^2$

$={PR}^2+{(RM+MQ)}^2$

$={PR}^2+{(RM+RM)}^2$ $(\because RM=MQ)$

$={PR}^2+{4RM}^2⟶\left(2\right)$

From $\left(1\right)$,

${RM}^2={PM}^2-{PR}^2⟶\left(3\right)$

Putting $\left(3\right)$ in $\left(2\right)$ we get

${PQ}^2={PR}^2+4({PM}^2-{PR}^2)$

${PQ}^2={PR}^2+{4PM}^2-{4PR}^2$

${PQ}^2={4PM}^2-3{PR}^2$

Hence proved.