<s>[INST] How did they get this equation comparing three ratios? I was reading from an old maths textbook. It was giving some examples on how to solve ratios. I stumbled upon this example and felt perplexed after reading only part of it.
We're given this equation.
$$\frac{x}{l(mb+nc-la)} = \frac{y}{m(nc+la-mb)} = \frac{z}{n(la + mb - nc)}$$
And asked to prove that
$$\frac{l}{x(by + cz - ax)} = \frac{m}{y(cz+ax-by)} = \frac{n}{z(ax + by -cz)}$$
He starts by doing this:
$$\frac{\frac{x}{l}}{mb + nc - la} = \frac{\frac{y}{m}}{nc + la - mb} = \frac{\frac{z}{n}}{la + mb -nc}$$
Which I understand. Then, he goes on to say this:

We have $$\frac{\frac{x}{l}}{mb + nc - la} = \frac{\frac{y}{m}}{nc + la - mb} = \frac{\frac{z}{n}}{la + mb -nc}$$
$$= \frac{\frac{y}{m} + \frac{z}{n}}{2la}$$
These are similar expressions.
$$\therefore  \frac{ny + mz}{a} = \frac{lz + nx}{b} = \frac{mx + ly}{c}$$

This is the portion of the proof that I don't understand. How did he go from
$= \frac{\frac{y}{m} + \frac{z}{n}}{2la}$ to $\frac{ny + mz}{a} = \frac{lz + nx}{b} = \frac{mx + ly}{c}$? And, also, what does he mean by these are "similar expressions."
The textbook I'm reading is called Higher Algebra a Sequel to Elementary Algebra for Schools by Henry Sinclair and Samuel Ratcliff Knight.
Thanks for the help.
 [/INST] The trick is that
$$\frac{a}{b}=\frac{c}{d} \implies \frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}$$
$$A=\frac{\frac{x}{l}}{mb + nc - la} = \frac{\frac{y}{m}}{nc + la - mb} = \frac{\frac{z}{n}}{la + mb -nc}$$
Use each two terms
$$A = \frac{\frac{x}{l} + \frac{y}{m}}{2nc} = \frac{\frac{x}{l} + \frac{z}{n}}{2mb} = \frac{\frac{y}{m} + \frac{z}{n}}{2la}$$
Then you arrive naturally at the end.
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