In $\Delta ABC$,$m\angle B = 90^\circ$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. If $AM = 4CM$,prove that $AB = 2BC$.

(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 $\Delta AMB \sim \Delta BMC$.
Therefore,the ratio of their corresponding sides is equal: $\frac{AB}{BC} = \frac{AM}{BM} = \frac{BM}{CM}$.
From $\frac{AB}{BC} = \frac{BM}{CM}$,we get $BM = \frac{AB \cdot CM}{BC}$.
Also,from $\Delta AMB \sim \Delta ABC$,we have $\frac{AB}{AC} = \frac{AM}{AB}$,which implies $AB^2 = AM \cdot AC$.
Similarly,from $\Delta BMC \sim \Delta ABC$,we have $\frac{BC}{AC} = \frac{CM}{BC}$,which implies $BC^2 = CM \cdot AC$.
Dividing the two equations: $\frac{AB^2}{BC^2} = \frac{AM \cdot AC}{CM \cdot AC} = \frac{AM}{CM}$.
Given $AM = 4CM$,we substitute this into the ratio: $\frac{AB^2}{BC^2} = \frac{4CM}{CM} = 4$.
Taking the square root of both sides: $\frac{AB}{BC} = \sqrt{4} = 2$.
Thus,$AB = 2BC$.