How does this line came? :)  
Today I started to learn algebra with my own. And the first chapter which I'm learning is Ratio. I'm kind of confuse in an example, in all the lines starting with (*). Until first three lines of the solution I understand but I didn't get how 
$$\frac{ny + mz}a = \frac{lz + nx}b = \frac{mx + ly}c$$ came. 
Please help and thank you in advance.
And sorry for my bad English. 

Example 2. If $\frac{x}{l(mb+nc-la)}=\frac{y}{m(nc+la-mb)}=\frac{z}{n(la+mb-nc)}$, prove that $\frac{l}{x(by+cz-ax)}=\frac{m}{y(cz+ax-by)}=\frac{n}{z(ax+by-cz)}$.
We have 
  \begin{align} \frac{\frac xl}{mb+nc-la} &= \frac{\frac ym}{nc+la-mb} = \frac{\frac zn}{la+mb-nc} \\ &= \frac{\frac ym + \frac zn}{2la} \\ &= \text{two similar expressions};   \end{align}
$$\therefore \quad \frac{ny+mz}a = \frac{lz+nx}b = \frac{mx+ly}c. \tag{*}$$
Multiply the first of these fractions above and below by $x$, the second by $y$, and the third by $z$; then
  \begin{align} \frac{nxy+mxz}{ax} &= \frac{lyz+nxy}{by} = \frac{mxz+lyz}{cz} \\ &= \frac{2lyz}{by+cz-ax} \\ &= \text{two similar expressions}; \end{align}
$$\therefore \quad \frac{l}{x(by+cz-ax)}=\frac{m}{y(cz+ax-by)}=\frac{n}{z(ax+by-cz)}$$

 A: If we have:$$\frac{a}{b}=\frac{c}{d}$$then this implies:$$\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\tag{1}$$You have:$$\frac{\frac{x}{l}}{mb+nc-la}=\frac{\frac{y}{m}}{nc+la-mb}=\frac{\frac{z}{n}}{la+mb-nc}$$Since these 3 terms are all equal, we can apply (1) to any two of them. So, lets start by applying (1) to the first 2 terms to get:$$\frac{\frac{x}{l}+\frac{y}{m}}{mb+nc-la+nc+la-mb}=\frac{\frac{x}{l}+\frac{y}{m}}{2nc}$$Similarly, applying (1) to the 2nd the 3rd terms gives:$$\frac{\frac{y}{m}+\frac{z}{n}}{nc+la-mb+la+mb-nc}=\frac{\frac{y}{m}+\frac{z}{n}}{2la}$$And finally, applying (1) to the 1st the 3rd terms gives:$$\frac{\frac{x}{l}+\frac{z}{n}}{mb+nc-la+la+mb-nc}=\frac{\frac{x}{l}+\frac{z}{n}}{2mb}$$All 3 of these must be equal to one another, therefore:$$\frac{\frac{x}{l}+\frac{y}{m}}{2nc}=\frac{\frac{y}{m}+\frac{z}{n}}{2la}=\frac{\frac{x}{l}+\frac{z}{n}}{2mb}$$We can multiply everything by 2 to get:$$\frac{\frac{x}{l}+\frac{y}{m}}{nc}=\frac{\frac{y}{m}+\frac{z}{n}}{la}=\frac{\frac{x}{l}+\frac{z}{n}}{mb}$$$$\therefore\frac{mx+ly}{lmnc}=\frac{ny+mz}{lmna}=\frac{nx+lz}{lmnb}$$Finally we multiply everything by $lmn$ to get:$$\frac{mx+ly}{c}=\frac{ny+mz}{a}=\frac{nx+lz}{b}$$
