$PQR$ is a triangle right-angled at $P$ and $M$ is a point on $QR$ such that $PM \perp QR$. Show that $PM^{2} = QM \cdot MR$.

(N/A) Let $\angle MPR = x$.
In $\Delta MPR$,
$\angle MRP = 180^{\circ} - 90^{\circ} - x = 90^{\circ} - x$.
Similarly,in $\Delta MPQ$,
$\angle MPQ = 90^{\circ} - \angle MPR = 90^{\circ} - x$.
In $\Delta MPQ$,
$\angle MQP = 180^{\circ} - 90^{\circ} - (90^{\circ} - x) = x$.
Now,in $\Delta QMP$ and $\Delta PMR$:
$\angle MPQ = \angle MRP = 90^{\circ} - x$
$\angle MQP = \angle MPR = x$
$\angle PMQ = \angle RMP = 90^{\circ}$
Therefore,$\Delta QMP \sim \Delta PMR$ (by $AA$ similarity criterion).
Since the triangles are similar,the ratios of their corresponding sides are equal:
$\frac{QM}{PM} = \frac{PM}{MR}$
$\Rightarrow PM^{2} = QM \cdot MR$.