A property of proportions: if $a/b=c/d$, then $(ma+nb)/(pa+qb)$ is equal to $ (mc+nd)/(pc+qd)$

Question

If $\large\frac{a}{b}=\frac{c}{d}$ how we can obtain $\displaystyle{\frac{ma+nb}{pa+qb}=\frac{mc+nd}{pc+qd}}$?

I can get $\large\frac{ma}{qb}=\frac{mc}{qd}$ and $\large\frac{nb}{pa}=\frac{nd}{pc}$ , now

If we have $\large\frac{a}{b}=\frac{c}{d}$ and $\large\frac{e}{f}=\frac{g}{h}$ is true that $\large\frac{a+e}{b+f}=\frac{c+g}{d+h}$ , I tried a particular solution and is not true at least for most cases,but for the above expression is true so how do I get there?can you help me out?

Answer 1

Hint:$$\frac{ma+nb}{pa+qb}=\frac{m\frac{a}{b}+n}{p\frac{a}{b}+q}$$

Answer 2

Follows immediately by $\rm\color{#c00}{M} =$ mediant, or by denominator linearity ("Or" below)

$x = \dfrac{\color{c00}a}{\color{c00}c}= \dfrac{\color{c00}b}{\color{c00}d}\,\Rightarrow\, x = \dfrac{m\color{c00}a}{m\color{c00}c} = \dfrac{n\color{c00}b}{n\color{c00}d}\,\overset{\rm\color{#c00}{M}}\Rightarrow\, x = \dfrac{ma\!+\!nb}{mc\!+\!nd}\ \left[ = \dfrac{pa\!+\!qb}{pc\!+\!qd}\,\ \rm similarly\right]$

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$\begin{align} {\bf Or}\qquad\qquad m\, &\left[cx = a\right]\\ +\ \ n\,&\left[dx = b\right]\\ \hline \Longrightarrow \ \ (mc\!+\!nd)\,x\!\!\!\!\!\!\!\! &\quad\ \ = ma\!+\!nd \end{align}$