In the given figure, it is given that ∠ABD = ∠CDB = ∠PQB = 90°. If AB = x units, CD = y units and PQ = z units, prove that $\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{z}$.

In ΔPQD and ΔABD,

∠ABD = ∠PQD = 90° [From figure]

∠ADB = ∠PDQ [Common angles]

∴ ΔPQD ∼ ΔABD (By A.A. axiom)

Corresponding sides of similar triangles are proportional.

$\Rightarrow \dfrac{PQ}{AB} = \dfrac{QD}{BD} \\[1em] \Rightarrow \dfrac{z}{x} = \dfrac{QD}{BD} \text{ …..(1)}$

In ΔPQB and ΔCDB,

∠CDB = ∠PQB = 90° [Given]

∠CBD = ∠PBQ [Common angles]

∴ ΔPQB ∼ ΔCDB (By A.A. axiom)

Corresponding sides of similar triangles are proportional.

$\Rightarrow \dfrac{PQ}{CD} = \dfrac{BQ}{BD} \\[1em] \Rightarrow \dfrac{z}{y} = \dfrac{BQ}{BD} \text{ ….(2)}$

Add equations (1) and (2) we get,

$\Rightarrow \dfrac{z}{x} + \dfrac{z}{y} = \dfrac{QD}{BD} + \dfrac{BQ}{BD} \\[1em] \Rightarrow z \Big(\dfrac{1}{x} + \dfrac{1}{y}\Big) = \dfrac{BD}{BD} \\[1em] \Rightarrow z \Big(\dfrac{1}{x} + \dfrac{1}{y}\Big) = 1 \\[1em] \Rightarrow \dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{z}.$

Hence, proved that $\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{z}$.