How did they get this equation comparing three ratios?

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.

Write $$t=\frac{\frac xl}{mb+nc-la}=\frac{\frac ym}{nc+la-mb}$$ then $$\frac xl=t(mb+nc-la)$$ and $$\frac ym=t(nc+la-mb).$$ Adding, $$\frac xl+\frac ym=2tnc$$ etc.

It is known that if $\frac {a_1}{b_1} = \frac {a_2}{b_2} = \frac {a_3}{b_3}=t$ then $\frac {a_1 + a_2}{b_1 + b_2}=\frac {a_1 + a_3}{b_1 + b_3}=\frac {a_2 + a_3}{b_2 + b_3}=\frac {a_1 + a_2 + a_3}{b_1 + b_2 + b_3}=t$