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

(A) In $\Delta ABC$,$\angle B = 90^{\circ}$ and $BE \perp AC$.
By the property of similarity in right-angled triangles,if an altitude is drawn to the hypotenuse,the triangles on both sides of the altitude are similar to the whole triangle and to each other.
$1$. $\Delta ABE \sim \Delta ABC$ (by $AA$ similarity).
Therefore,$\frac{AB}{AC} = \frac{AE}{AB}$,which implies $AB^2 = AE \cdot AC$ --- $(1)$.
$2$. $\Delta CBE \sim \Delta ABC$ (by $AA$ similarity).
Therefore,$\frac{BC}{AC} = \frac{CE}{BC}$,which implies $BC^2 = CE \cdot AC$ --- $(2)$.
Dividing equation $(1)$ by equation $(2)$:
$\frac{AB^2}{BC^2} = \frac{AE \cdot AC}{CE \cdot AC}$.
Canceling $AC$ from the numerator and denominator,we get:
$\frac{AB^2}{BC^2} = \frac{AE}{CE}$.
Hence proved.