In $\Delta PQR$,$m \angle Q = 90^{\circ}$ and $\overline{QD}$ is an altitude to the hypotenuse $PR$. If $PD = 25 DR$,prove that $PQ = 5 QR$.

(N/A) In $\Delta PQR$,$\angle Q = 90^{\circ}$ and $QD \perp PR$.
By the property of geometric mean in a right-angled triangle,we have $\Delta PDQ \sim \Delta QDR$.
From the similarity of triangles,the ratio of corresponding sides is equal: $\frac{PQ}{QR} = \frac{PD}{QD} = \frac{QD}{DR}$.
From $\frac{PQ}{QR} = \frac{QD}{DR}$,we get $PQ^2 = QR^2 \cdot \frac{PD}{DR}$.
Given $PD = 25 DR$,so $\frac{PD}{DR} = 25$.
Substituting this value,$PQ^2 = QR^2 \cdot 25$.
Taking the square root of both sides,we get $PQ = 5 QR$.