Question

 If a : b = c : d and e : f = g : h, prove that ae + bf : ae – bf = cg + dh : cg –dh

Answer 1

Hint: Here we have to multiply both the equations by dividendo and componendo. As per this rule, if $\dfrac{a}{b} = \dfrac{c}{d}$, Then this can be expressed as $\dfrac{{a + b}}{{a - b}} = \dfrac{{c + d}}{{c - d}}$

Complete step-by-step answer:

Given, $\dfrac{a}{b} = \dfrac{c}{d}$ —--- (1)

$\dfrac{e}{f} = \dfrac{g}{h}$ —-- (2)

Multiplying both the equation (1) and (2)

$\dfrac{{ae}}{{bf}} = \dfrac{{cg}}{{dh}}$

By dividendo and componendo

$\dfrac{{ae + bf}}{{ae - bf}} = \dfrac{{cg + dh}}{{cg - dh}}$

Note:

When faced with this type of problem, first convert the given form to a rational form that multiplies both equations and then apply the dividendo and componendo to obtain the final answer.