In $\mathrm{△}\mathrm{PQR},\mathrm{QM}\perp \mathrm{PR}$ and ${\mathrm{PR}}^2-{\mathrm{PQ}}^2={\mathrm{QR}}^2$. Then ${\mathrm{QM}}^2=$

In $\mathrm{\Delta PQR}$, we have
$$\begin{gathered} {\mathrm{PR}}^2-{\mathrm{PQ}}^2={\mathrm{QR}}^2 \\ \therefore {\mathrm{PR}}^2={\mathrm{PQ}}^2+{\mathrm{QR}}^2 \end{gathered}$$
$\Rightarrow \mathrm{\Delta PQR}$ is a right triangle right – angles at Q
$\Rightarrow \angle 2+\angle 3=90°$

Also, $\mathrm{\angle }1+\mathrm{\angle }2={90}^{\circ }\left[\because \mathrm{\angle }\mathrm{PMQ}={90}^{\circ }\right]$
$\therefore \angle 1=\angle 3$
Similarly, we have
$\angle 2=\angle 4$
Thus, in $\mathrm{\Delta }\text{'}\mathrm{s}$ PMQ and QMR, we have
$\angle 1=\angle 3 \mathrm{and} \angle 2=\angle 4$
So, by AA- criterion for similarity, we obtain
$\begin{array}{l}\mathrm{△}\mathrm{PMQ}~\mathrm{\Delta QMR}\\ \Rightarrow \frac{\mathrm{PM}}{\mathrm{QM}}=\frac{\mathrm{MQ}}{\mathrm{MR}}\\ \Rightarrow {\mathrm{QM}}^2=\mathrm{PM}\times \mathrm{MR}\end{array}$