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

(N/A) In $\Delta ABC$,$\angle B = 90^{\circ}$ and $\overline{BM} \perp \overline{AC}$.
By the property of geometric mean in a right-angled triangle,we have $BM^2 = AM \cdot CM$.
Given that $BM = 2 CM$,substitute this into the equation:
$(2 CM)^2 = AM \cdot CM$
$4 CM^2 = AM \cdot CM$
Since $CM \neq 0$,we can divide both sides by $CM$ to get $AM = 4 CM$.
Now,the hypotenuse $AC$ is the sum of segments $AM$ and $CM$:
$AC = AM + CM$
$AC = 4 CM + CM$
$AC = 5 CM$.
Hence,the result is proved.