In $\Delta ABC$,$m\angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. Prove that $\frac{1}{BM^2} = \frac{1}{AB^2} + \frac{1}{BC^2}$.

(N/A) In $\Delta ABC$,$m\angle B = 90^{\circ}$ and $\overline{BM} \perp \overline{AC}$.
By the area of a right-angled triangle,the area of $\Delta ABC$ can be expressed in two ways:
Area $= \frac{1}{2} \times AB \times BC$ (using legs as base and height)
Area $= \frac{1}{2} \times AC \times BM$ (using hypotenuse as base and altitude as height)
Equating the two expressions: $\frac{1}{2} \times AB \times BC = \frac{1}{2} \times AC \times BM \implies AB \times BC = AC \times BM$.
Squaring both sides: $AB^2 \times BC^2 = AC^2 \times BM^2$.
Since $\Delta ABC$ is a right-angled triangle,by the Pythagorean theorem: $AC^2 = AB^2 + BC^2$.
Substituting $AC^2$ into the equation: $AB^2 \times BC^2 = (AB^2 + BC^2) \times BM^2$.
Rearranging for $BM^2$: $BM^2 = \frac{AB^2 \times BC^2}{AB^2 + BC^2}$.
Taking the reciprocal of both sides: $\frac{1}{BM^2} = \frac{AB^2 + BC^2}{AB^2 \times BC^2}$.
Splitting the fraction: $\frac{1}{BM^2} = \frac{AB^2}{AB^2 \times BC^2} + \frac{BC^2}{AB^2 \times BC^2}$.
Simplifying: $\frac{1}{BM^2} = \frac{1}{BC^2} + \frac{1}{AB^2}$.