$\Delta ABC$ is a triangle where $\angle C=90^{\circ }$. If $p$ is the length of the perpendicular from $C$ to $AB$, then $\frac{1}{B{C}^2}+\frac{1}{A{C}^2}=$

Let the lengths of the sides opposite to angles $A,B,C$ be $a,b,c$ respectively. So, $BC=a$, $AC=b$, and $AB=c$.
Since $\angle C=90^{\circ }$, by the Pythagorean theorem:

$$c^2=a^2+b^2⟹A{B}^2=B{C}^2+A{C}^2$$

The area of $\Delta ABC$ can be calculated in two ways:

1. Using base $BC$ and height $AC$:

$$\text{Area}=\frac{1}{2}\times a\times b$$

2. Using base $AB$ and perpendicular height $p$:

$$\text{Area}=\frac{1}{2}\times c\times p$$

Equating the two expressions for area:

$$\frac{1}{2}ab=\frac{1}{2}cp⟹ab=cp⟹c=\frac{ab}{p}$$

Squaring both sides:

$$c^2=\frac{a^2{b}^2}{p^2}$$

Substitute $c^2=a^2+b^2$:

$$a^2+b^2=\frac{a^2{b}^2}{p^2}$$

Divide both sides by $a^2{b}^2$:

$$\frac{a^2}{a^2{b}^2}+\frac{b^2}{a^2{b}^2}=\frac{1}{p^2}$$

$$\frac{1}{b^2}+\frac{1}{a^2}=\frac{1}{p^2}$$

Replacing $a$ and $b$ with $BC$ and $AC$:

$$\frac{1}{B{C}^2}+\frac{1}{A{C}^2}=\frac{1}{p^2}$$