In $\Delta ABC$,$m\angle B = 90^{\circ}$ and $\overline{BD}$ is an altitude to the hypotenuse $\overline{AC}$. If $AC = 5 CD$,prove that $BD = 2 CD$.

(N/A) In $\Delta ABC$,$\angle B = 90^{\circ}$ and $\overline{BD} \perp \overline{AC}$.
By the property of geometric mean in a right-angled triangle,we have $BD^2 = AD \cdot CD$.
Given $AC = 5 CD$. Since $AC = AD + CD$,we have $AD + CD = 5 CD$,which implies $AD = 4 CD$.
Substituting $AD = 4 CD$ into the geometric mean equation:
$BD^2 = (4 CD) \cdot CD$
$BD^2 = 4 CD^2$
Taking the square root of both sides:
$BD = \sqrt{4 CD^2} = 2 CD$.
Thus,it is proved that $BD = 2 CD$.