<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] To prove your requested expression (note there are limitations as Gerry Myerson's question comment explains, e.g., $b \neq 0$, $d \neq 0$, $b + d \neq 0$, $b - d \neq 0$, etc.), cross-multiply and use other manipulations to get
$$\begin{equation}\begin{aligned}
\frac{a}{b} & = \frac{c}{d} \\
ad & = bc \\
ad + cd & = bc + cd \\
d(a + c) & = c(b + d) \\
\frac{a + c}{b + d} & = \frac{c}{d}
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
You can similarly prove that $\frac{c}{d} = \frac{a - c}{b - d}$. Note, however, you can prove the $2$ requested identities more simply as shown below. The Law of sines gives
$$\frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)} \tag{2}\label{eq2A}$$
Cross-multiplying the left & right parts gives
$$\begin{equation}\begin{aligned}
a\sin(C) & = c\sin(A) \\
\frac{a}{c} & = \frac{\sin(A)}{\sin(C)}
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
Similarly, using the middle & right parts in \eqref{eq2A} gives
$$\frac{b}{c} = \frac{\sin(B)}{\sin(C)} \tag{4}\label{eq4A}$$
Adding \eqref{eq3A} and \eqref{eq4A} gives
$$\frac{a+b}{c}=\frac{\sin(A) +\sin(B)}{\sin(C)} \tag{5}\label{eq5A}$$
while \eqref{eq3A} minus \eqref{eq4A} gives
$$\frac{a-b}{c}=\frac{\sin(A) -\sin(B)}{\sin(C)} \tag{6}\label{eq6A}$$
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