Question

this is my challenge=
if u give right answer than i will follow u
only intelligents give answer please ​

Answer 1

The correct answer is  $\mathbf{c.\ \ \ \dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}}$

This is a rule known as "rule of componendo and dividendo".

Let me prove it.

We have  $\dfrac{a}{b}\ =\ \dfrac{c}{d}.$

By cross multiplication, we get  $ad=bc.$

[This gives option (b) incorrect unless c = d.]

So we can have,

$ad-bc=bc-ad=0\\ \\ \text{OR}\ \ \ ad-bc=-ad+bc$

Add  $ac-bd$  to both sides.

$ac+ad-bc-bd=ac-ad+bc-bd\\ \\ \implies\ a(c+d)-b(c+d)=a(c-d)+b(c-d)\\ \\ \implies\ (a-b)(c+d)=(a+b)(c-d)\\ \\ \implies\ \dfrac{a-b}{a+b}=\dfrac{c-d}{c+d}$

Okay, proved!

The rule of componendo and dividendo is just the substitution of the rule of componendo and the rule of dividendo.

Rule of componendo states that  $\dfrac{a}{b}=\dfrac{c}{d}\ \iff\ \dfrac{a+b}{b}=\dfrac{c+d}{d}.$

Rule of dividendo states that  $\dfrac{a}{b}=\dfrac{c}{d}\ \iff\ \dfrac{a-b}{b}=\dfrac{c-d}{d}$

It is always true that in two equivalent fractions, the ratio between the numerators and that between the denominators are proportional.

Means,  $\dfrac{a}{b}=\dfrac{c}{d}\ \iff\ \dfrac{a}{c}=\dfrac{b}{d}$

Because it also implies  $ad=bc.$

If  $\dfrac{a}{b}=\dfrac{c}{d},$  this won't imply  $ab=cd,$  but it will be true that  $ab$  is a multiple or sub multiple of  $cd.$

$ab=\left(\dfrac{a}{c}\right)^2cd\quad\quad\text{OR}\quad\quad cd=\left(\dfrac{c}{a}\right)^2ab\\ \\ \\ \text{Since}\ \ \dfrac{a}{c}=\dfrac{b}{d}\ \iff\ \dfrac{c}{a}=\dfrac{d}{b},\\ \\ \\ ab=\left(\dfrac{b}{d}\right)^2cd\quad\quad\text{OR}\quad\quad cd=\left(\dfrac{d}{b}\right)^2ab$

So we get option (a) incorrect.

Option (a) is always incorrect, that,

$\dfrac{a}{b}=\dfrac{c}{d}\ \ \ \ \ \not\!\!\!\!\!\iff\ \ \ \ \ \dfrac{a}{d}=\dfrac{c}{b}$

Therefore,  $\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\ \iff\ \dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\ \ \ \ \ \not\!\!\!\!\!\iff\ \ \ \ \ \dfrac{a+b}{c-d}=\dfrac{a-b}{c+d}$

So option (d) is incorrect too!