# 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.

$$\frac{x}{l(mb+nc-la)} = \frac{y}{m(nc+la-mb)} = \frac{z}{n(la + mb - nc)}$$

$$\frac{l}{x(by + cz - ax)} = \frac{m}{y(cz+ax-by)} = \frac{n}{z(ax + by -cz)}$$

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.

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.

• $\tfrac12$, $\tfrac34$ and $\tfrac{1+3}{2+4}$ are all distinct, so what did you actually mean to write for your first line? – David Richerby Jan 6 '18 at 20:02
• @Davis Richerby the first two fractions have to be equal – DreamConspiracy Jan 6 '18 at 20:04

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

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.

• When I sum $t(mb + nc - la)$ and $t(nc + la - mb)$ I obtain the result $2tnc$. How is this equivalent to $2anc$? – ssharma Jan 6 '18 at 18:15