What are some good questions for this trick, if $\frac{a}{b}=\frac{c}{d}=\frac{e}{f}=\dots=\alpha$ then $\alpha=\frac{a+c+e+...}{b+d+f+...}$? I need some good algebra questions that are applications of this trick, often in a non obvious and elegant way:   $$\text{If } \frac{a}{b}=\frac{c}{d}=\frac{e}{f}=\dots=\alpha \text{ then } \alpha=\frac{a+c+e+...}{b+d+f+...}$$
 A: *

*If $\displaystyle\frac a{b+c}=\frac b{c+a}=\frac c{a+b};$  prove that each ratio $\displaystyle=\frac12$ if $\displaystyle a+b+c\ne0$

*If  $\displaystyle\frac{a-b}{x^2}=\frac{b-c}{y^2}=\frac{c-a}{z^2}$ prove that $\displaystyle a=b=c$

*If  $\displaystyle\frac{a-b}{a^2+ab+b^2}=\frac{b-c}{b^2+bc+c^2}=\frac{c-a}{c^2+ca+a^2}$ prove that $\displaystyle a=b=c$ (for a special condition)

*If  $\displaystyle\frac{a+b}{a^2+ab+b^2}=\frac{b+c}{b^2+bc+c^2}$ and $a\ne b\ne c$ prove that each ratio $=\displaystyle\frac{c+a}{c^2+ca+a^2}$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$\begin{align}
\color{#c00000}{\large\alpha}&={a \over b}=\color{#c00000}{\large{c \over d}}\ \imp\ {a \over c}={b \over d}\ \imp\ {a \over c} + 1 ={b \over d} + 1\
\imp\ {a + c \over c}={b + d \over d}\ \imp\
{a + c \over b + d} = \color{#c00000}{\large{c \over d}} 
\end{align}

$$
\mbox{Then,}\quad  {a + c \over b + d} = \alpha
$$

Now, we have
$$
{a + c \over b + d}={e \over f}=\cdots=\alpha
$$

We can repeat the above procedure as needed. The general result will follow.

