In $\Delta PQR$,$m \angle Q = 90^{\circ}$ and $\overline{QD}$ is an altitude to the hypotenuse $\overline{PR}$. If $PQ = 3QR$,prove that $PD = 9RD$.

(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,$\Delta PDQ \sim \Delta PQR$ and $\Delta RDQ \sim \Delta RQP$.
$3$. Specifically,$\Delta PDQ \sim \Delta QDR$ is not directly applicable,but we use the property that in a right triangle,the altitude to the hypotenuse divides the triangle into two triangles similar to the original and to each other.
$4$. Thus,$\Delta PDQ \sim \Delta QDR$.
$5$. From the similarity $\Delta PDQ \sim \Delta QDR$,we have the ratio of corresponding sides: $\frac{PD}{QD} = \frac{QD}{RD} = \frac{PQ}{QR}$.
$6$. We are given $PQ = 3QR$,which implies $\frac{PQ}{QR} = 3$.
$7$. From $\frac{PD}{QD} = \frac{PQ}{QR}$,we get $PD = 3QD$.
$8$. From $\frac{QD}{RD} = \frac{PQ}{QR}$,we get $QD = 3RD$.
$9$. Substituting $QD = 3RD$ into $PD = 3QD$,we get $PD = 3(3RD) = 9RD$.
$10$. Hence,$PD = 9RD$ is proved.