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

(N/A) Given: In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM} \perp \overline{AC}$.
To prove: $\frac{AB^2}{BC^2} = \frac{AM}{CM}$.
Proof: In $\Delta ABC$,since $\overline{BM}$ is the altitude to the hypotenuse,we have two similar triangles:
$1$. $\Delta AMB \sim \Delta ABC$,which implies $\frac{AB}{AC} = \frac{AM}{AB}$,so $AB^2 = AM \times AC$.
$2$. $\Delta BMC \sim \Delta ABC$,which implies $\frac{BC}{AC} = \frac{CM}{BC}$,so $BC^2 = CM \times AC$.
Dividing the two equations:
$\frac{AB^2}{BC^2} = \frac{AM \times AC}{CM \times AC}$
Therefore,$\frac{AB^2}{BC^2} = \frac{AM}{CM}$.