In $\Delta PQR$,$m\angle Q = 90^{\circ}$ and $\overline{PM}$ is a median. Prove that $PR^{2} = PM^{2} + 3RM^{2}$.

(N/A) $1$. In $\Delta PQR$,$\angle Q = 90^{\circ}$. By Pythagoras theorem,$PR^{2} = PQ^{2} + QR^{2}$.
$2$. Since $\overline{PM}$ is a median to side $\overline{QR}$,$M$ is the midpoint of $\overline{QR}$. Therefore,$QM = RM = \frac{1}{2} QR$,which implies $QR = 2RM$.
$3$. In right-angled $\Delta PQM$,by Pythagoras theorem,$PM^{2} = PQ^{2} + QM^{2}$. Thus,$PQ^{2} = PM^{2} - QM^{2}$.
$4$. Substitute $PQ^{2} = PM^{2} - QM^{2}$ and $QR = 2RM$ into the first equation: $PR^{2} = (PM^{2} - QM^{2}) + (2RM)^{2}$.
$5$. Since $QM = RM$,substitute $QM$ with $RM$: $PR^{2} = PM^{2} - RM^{2} + 4RM^{2}$.
$6$. Simplifying the expression,we get $PR^{2} = PM^{2} + 3RM^{2}$. Hence proved.