In a right-angled ΔABC, right-angled at A, if AD ⟂ BC such that AD = p, BC = a, CA = b and AB = c, then:

1. $\dfrac{p}{a} = \dfrac{p}{b}$

2. p² = b² + c²

3. p² = b² c²

4. $\dfrac{1}{p^{2}} = \dfrac{1}{b^{2}} + \dfrac{1}{c^{2}}$

Given,

Area of a right angled triangle = $\dfrac{1}{2} \times \text{ base } \times \text{ height }$

Area of triangle ABC = $\dfrac{1}{2} \times AB \times AC$

Area of triangle ABC = $\dfrac{1}{2} \times bc$ …….(1)

Also,

Area of triangle ABC = $\dfrac{1}{2} \times BC \times AD$

Area of triangle ABC = $\dfrac{1}{2} \times ap$ …..(2)

From equation (1) and (2), we get :

$\dfrac{1}{2}$ bc = $\dfrac{1}{2}$ ap

bc = ap

a = $\dfrac{bc}{p}$

Applying Pythogoras theorem to right-angled triangle ABC:

$\Rightarrow BC^2 = AC^2 + AB^2 \\[1em] \Rightarrow a^2 = b^2 + c^2 \\[1em] \Rightarrow \Big(\dfrac{bc}{p}\Big)^2 = b^2 + c^2 \\[1em] \Rightarrow \dfrac{b^2c^2}{p^2} = b^2 + c^2 \\[1em] \Rightarrow \dfrac{1}{p^2} = \dfrac{b^2 + c^2}{b^2c^2} \\[1em] \Rightarrow \dfrac{1}{p^2} = \dfrac{b^2}{b^2c^2} + \dfrac{c^2}{b^2c^2}\\[1em] \Rightarrow \dfrac{1}{p^2} = \dfrac{1}{c^2} + \dfrac{1}{b^2} .$

Hence, option 4 is the correct option.