In a $\triangle PQR$,$PR^{2} - PQ^{2} = QR^{2}$ and $M$ is a point on side $PR$ such that $QM \perp PR$. Prove that $QM^{2} = PM \times MR$.

(N/A) Given: In $\triangle PQR$,$PR^{2} - PQ^{2} = QR^{2}$ and $QM \perp PR$.
To prove: $QM^{2} = PM \times MR$.
Proof: Since $PR^{2} - PQ^{2} = QR^{2}$,we have $PR^{2} = PQ^{2} + QR^{2}$.
By the converse of the Pythagoras theorem,$\triangle PQR$ is a right-angled triangle at $Q$.
In $\triangle QMR$ and $\triangle PMQ$:
$\angle M = \angle M = 90^{\circ}$ (given $QM \perp PR$ and $\angle PQR = 90^{\circ}$ implies $\angle MQR + \angle MQP = 90^{\circ}$).
Since $\angle P + \angle R = 90^{\circ}$ and $\angle MQR + \angle R = 90^{\circ}$,we have $\angle MQR = \angle P$.
Similarly,$\angle MQP = \angle R$.
Thus,$\triangle QMR \sim \triangle PMQ$ by $AA$ similarity criterion.
Since the triangles are similar,the ratios of their corresponding sides are equal:
$\frac{QM}{PM} = \frac{MR}{QM}$.
Therefore,$QM^{2} = PM \times MR$. Hence proved.