If $\frac{a}{b}=\frac{c}{d}$, is it possible to write $\frac{b-a}{b+a} = \frac{c-d}{c+d}$?

Question

Consider, $$\frac{a}{b} = \frac{c}{d}$$ From this is it possible to write, $$\frac{b-a}{b+a} = \frac{c-d}{c+d}$$

Edit: It is how to get the second expression from the first

Answer 1

Divide numerator and denominator of the left hand side by $b$, and of the right hand side by $d$. Then $$\frac{b-a}{b+a}=\frac{1-a/b}{1+a/b}=\frac{1-c/d}{1+c/d}=\frac{d-c}{d+c}\ .$$ So, not quite what you asked but almost. Of course we have to assume that $b$ and $d$ and $b+a$ and $d+c$ are not zero.

Answer 2

I will use determinant to show this. Of course the benefit is that in doing so it will allow you to introduce linear algebra and geometry into understanding the equation that you want to prove in the picture.

Consider the matrix \begin{align} A= \begin{bmatrix} a & c\\ b & d \end{bmatrix}. \end{align} Notice that the first condition is equivalent to $\det A = 0$ with $c, d\ne 0$.

By row replacement, we have that \begin{align} A\sim \begin{bmatrix} a & c\\ b-a & d-c \end{bmatrix} \sim \begin{bmatrix} \frac{b+a}{2} & \frac{c+d}{2}\\ b-a & d-c \end{bmatrix}. \end{align} Since row replacement doesn't change the determinant, then it follows that \begin{align} \frac{b+a}{2}(d-c) = \frac{c+d}{2}(b-a). \end{align} This yields the desired result.