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

(N/A) Given: In $\Delta PQR$,$m \angle Q = 90^{\circ}$,$\overline{QD}$ is an altitude to the hypotenuse $PR$,and $PD = 9RD$.
To prove: $PQ = 3QR$.
Proof: In $\Delta PQR$,since $m \angle Q = 90^{\circ}$ and $\overline{QD} \perp \overline{PR}$,by the geometric mean theorem (or properties of similar triangles formed by an altitude to the hypotenuse):
$PQ^2 = PD \times PR$ and $QR^2 = RD \times PR$.
Dividing the two equations:
$\frac{PQ^2}{QR^2} = \frac{PD \times PR}{RD \times PR} = \frac{PD}{RD}$.
Substituting $PD = 9RD$ into the equation:
$\frac{PQ^2}{QR^2} = \frac{9RD}{RD} = 9$.
Taking the square root on both sides:
$\frac{PQ}{QR} = 3$.
Therefore,$PQ = 3QR$.