In $\Delta PQR$,$\overline{PM}$ is an altitude. If $PM^2 = QM \times RM$,prove that $\Delta PQR$ is a right-angled triangle.

(N/A) Given: In $\Delta PQR$,$\overline{PM} \perp \overline{QR}$ and $PM^2 = QM \times RM$.
Step $1$: From the given equation,we have $\frac{PM}{QM} = \frac{RM}{PM}$.
Step $2$: In $\Delta PMQ$ and $\Delta RMP$,$\angle PMQ = \angle RMP = 90^\circ$ (since $\overline{PM}$ is an altitude).
Step $3$: By the $SAS$ similarity criterion,$\Delta PMQ \sim \Delta RMP$.
Step $4$: Since the triangles are similar,their corresponding angles are equal: $\angle QPM = \angle PRM$ and $\angle PQM = \angle RPM$.
Step $5$: In $\Delta PQR$,the sum of angles is $180^\circ$. Thus,$\angle P + \angle Q + \angle R = 180^\circ$.
Step $6$: Substituting the angles,we get $\angle QPM + \angle RPM + \angle Q + \angle R = 180^\circ$.
Step $7$: Since $\angle QPM = \angle R$ and $\angle RPM = \angle Q$,we have $\angle R + \angle Q + \angle Q + \angle R = 180^\circ$,which simplifies to $2(\angle Q + \angle R) = 180^\circ$,so $\angle Q + \angle R = 90^\circ$.
Step $8$: Therefore,$\angle P = 180^\circ - 90^\circ = 90^\circ$. Hence,$\Delta PQR$ is a right-angled triangle.