In the following figure, AB, CD and EF are perpendicular to the straight line BDF.

If AB = x and, CD = z unit and EF = y unit, prove that:

$\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{z}$

Let BD = a and DF = b.

In ΔFDC and ΔFBA,

∠FDC = ∠FBA [Both = 90°]

∠DFC = ∠BFA [Common angle]

∴ ∆FDC ~ ∆FBA [By AA]

Since, corresponding sides of similar triangle are proportional to each other.

$\Rightarrow \dfrac{DF}{BF} = \dfrac{DC}{AB} \\[1em] \Rightarrow \dfrac{DF}{BD + DF} = \dfrac{z}{x} \\[1em] \Rightarrow \dfrac{b}{a + b} = \dfrac{z}{x} ….. (1)$

In ΔBDC and ΔBFE,

∠BDC = ∠BFE [Both = 90°]

∠DBC = ∠FBE [Common angle]

∴ ∆BDC ~ ∆BFE [By AA]

Since, corresponding sides of similar triangle are proportional to each other.

$\Rightarrow \dfrac{BD}{BF} = \dfrac{CD}{EF} \\[1em] \Rightarrow \dfrac{BD}{BD + DF} = \dfrac{z}{y} \\[1em] \Rightarrow \dfrac{a}{a + b} = \dfrac{z}{y} ………… (2)$

Adding (1) and (2) we get :

$\Rightarrow \dfrac{b}{a + b} + \dfrac{a}{a + b} = \dfrac{z}{x} + \dfrac{z}{y} \\[1em] \Rightarrow \dfrac{a + b}{a + b} = z\Big(\dfrac{1}{x} + \dfrac{1}{y}\Big) \\[1em] \Rightarrow 1 = z\Big(\dfrac{1}{x} + \dfrac{1}{y}\Big) \\[1em] \Rightarrow \dfrac{1}{z} = \dfrac{1}{x} + \dfrac{1}{y}.$

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