Why does $\tfrac{a}{b} = \tfrac{c}{d} = \tfrac{e}{f} \rightarrow \tfrac{a}{b} = \tfrac{c-e}{d-f}$

Question

Why is it that

$$\frac{a}{b} = \frac{c}{d} = \frac{e}{f} \rightarrow \frac{a}{b} = \frac{c-e}{d-f}$$

It is used in the proof of Ceva's theorem. Thank you.

Answer 1

assuming that none of the denominators are zero : $$\frac{a}{b} = \frac{c}{d} \rightarrow ad=bc$$ $$\frac{a}{b} = \frac{e}{f} \rightarrow af=be$$ $$ad-af=bc-be$$ $$a(d-f)=b(c-e)\rightarrow \frac{a}{b} = \frac{c-e}{d-f}$$

Answer 2

Call, $\frac{a}{b}=\frac{c}{d}=\frac{e}{f}=k$, then $c=dk$ and $e=fk$, and $a=bk$, substitute this to $\frac{a}{b}$ and to$\frac{c-e}{d-f}$, and you get your equality.

Answer 3

WLOG, assume $a,b,c,d,e,f$ are not $0.$ Then from $$\frac{c}{d}=\frac{e}{f}$$ we get $$\frac{c}{e}=\frac{d}{f},$$ and thus $$\frac{c}{e}-1=\frac{d}{f}-1,$$ that is, $$\frac{c-e}{e}=\frac{d-f}{f},$$ which is equivalent to $$\frac{c-e}{d-f}=\frac{e}{f}.$$ Consequently, we have $$\frac{a}{b}=\frac{e}{f}=\frac{c-e}{d-f}.$$

Answer 4

$\begin{array}{rrrl}{\bf Hint} && d\, x\!\!\! &=&\!\!\! c\\ && f\, x \!\!\!&=&\!\!\! e\\ \Rightarrow &&(d\!-\!f)\, x\!\!\! &=&\!\!\! c\!-\!e\ \ \text{ by subtracting the prior equations.}\end{array}$

Geometrically: if vectors $(d,c),\,(f,e)$ have the same slope, then so too does their difference.