If a + c = be and $\dfrac{1}{b} + \dfrac{1}{d} = \dfrac{e}{c}$, prove that : a, b, c and d are in proportion.

Given,

a + c = be and $\dfrac{1}{b} + \dfrac{1}{d} = \dfrac{e}{c}$.

Solving,

a + c = be

Divide the equation by b,

$\Rightarrow \dfrac{a}{b} + \dfrac{c}{b}$ = e …………(1)

Now solving,

$\dfrac{1}{b} + \dfrac{1}{d} = \dfrac{e}{c}$

Multiplying the equation by c,

$\Rightarrow \dfrac{c}{b} + \dfrac{c}{d}$ = e

Putting value of e from equation (1) in above equation, we get :

$\Rightarrow \dfrac{c}{b} + \dfrac{c}{d} = \dfrac{a}{b} + \dfrac{c}{b} \\[1em] \Rightarrow \dfrac{c}{d} = \dfrac{a}{b}.$

Since, $\dfrac{a}{b} = \dfrac{c}{d}$, hence, a, b, c and d are in proportion.