<s>[INST] How to prove $\frac{a}{b}=\frac{c}{d} \implies \frac{a+c}{b+d}=\frac{a-c}{b-d}$? My question:

How do I prove the following property of ratio?
$$\frac{a}{b}=\frac{c}{d} \implies \frac{a+c}{b+d}=\frac{a-c}{b-d} \tag{1}$$
$a,b,c,d \in \mathbb{R} \backslash \{ 0 \}$

I want to use the result in argument below.


Use the sine rule to establish the following identities for triangles:
$$\frac{a+b}{c}=\frac{\sin(A) +\sin(B)}{\sin(C)}$$ and \begin{equation}\frac{a-b}{c}=\frac{\sin(A) -\sin(B)}{\sin(C)}\end{equation}

To prove both these identities the equations rearrange to the following
$$\frac{\sin(C)}{c}=\frac{\sin(A)+\sin(B)}{a+b}=\frac{\sin(A)-\sin(B)}{a-b} $$
From here, $(1)$ would complete the argument.
Thanks in advance
 [/INST] But why not use this simple solution:
$$
\frac{a}{b}=\frac{c}{d}=x \implies a=bx , c = dx
$$
$$
\begin{align}
\frac{a+c}{b+d}&=\frac{bx+dx}{b+d}\\
\\
&=x\\
\\
\frac{a-c}{b-d}&=\frac{bx-dx}{b-d}\\
\\
&=x
\end{align}
$$
For the desired result, use sine rule
$$
\begin{align}
a&=2R\sin{(A)}\\
b&=2R\sin{(B)}\\
c&=2R\sin{(C)}
\end{align}
$$
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