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Question
AnswersIf a : b = c : d and e : f = g : h, prove that ae + bf : ae – bf = cg + dh : cg –dh
Answer
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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, $\dfrac{a}{b} = \dfrac{c}{d}$ equation (1)
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:
Whenever you come to this type of problem first convert the given form to a rational form that multiplies both the equation and after applying the dividendo and componendo get the final answer.
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, $\dfrac{a}{b} = \dfrac{c}{d}$ equation (1)
$\dfrac{e}{f} = \dfrac{g}{h}$ equation (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:
Whenever you come to this type of problem first convert the given form to a rational form that multiplies both the equation and after applying the dividendo and componendo get the final answer.