In $\Delta PQR$,$m \angle Q = 90^{\circ}$ and $\overline{QD}$ is an altitude to the hypotenuse $\overline{PR}$. If $PQ = 4QR$,prove that $PD = 16RD$.

(N/A) $1$. In $\Delta PQR$,$\angle Q = 90^{\circ}$ and $\overline{QD} \perp \overline{PR}$.
$2$. By the property of similarity in right-angled triangles,$\Delta PDQ \sim \Delta QDR \sim \Delta PQR$.
$3$. From $\Delta PDQ \sim \Delta QDR$,we have the ratio of corresponding sides: $\frac{PD}{QD} = \frac{QD}{RD} = \frac{PQ}{QR}$.
$4$. From the ratio $\frac{PD}{QD} = \frac{PQ}{QR}$,we get $PD = QD \cdot \frac{PQ}{QR}$.
$5$. From the ratio $\frac{QD}{RD} = \frac{PQ}{QR}$,we get $QD = RD \cdot \frac{PQ}{QR}$.
$6$. Substituting $QD$ in the expression for $PD$: $PD = (RD \cdot \frac{PQ}{QR}) \cdot \frac{PQ}{QR} = RD \cdot (\frac{PQ}{QR})^2$.
$7$. Given $PQ = 4QR$,so $\frac{PQ}{QR} = 4$.
$8$. Therefore,$PD = RD \cdot (4)^2 = 16RD$.