Answer: Using Pythagoras theorem in \(\triangle PRQ\) and \(\triangle PRM\) with QR = 2RM gives \(PQ^2 = 4PM^2 - 3PR^2\).
M is the midpoint of QR, so QR = 2RM, hence \(QR^2 = 4RM^2\).
In \(\triangle PRQ\), \(\angle PRQ = 90^\circ\):
\[PQ^2 = PR^2 + QR^2 = PR^2 + 4RM^2\quad \text{(I)}\]In \(\triangle PRM\), \(\angle PRM = 90^\circ\):
\[PM^2 = PR^2 + RM^2\] \[RM^2 = PM^2 - PR^2\quad \text{(II)}\]Substituting (II) in (I):
\[PQ^2 = PR^2 + 4PM^2 - 4PR^2\] \[PQ^2 = 4PM^2 - 3PR^2\]Hence proved.