Why does $\tfrac{a}{b} = \tfrac{c}{d} = \tfrac{e}{f} \rightarrow \tfrac{a}{b} = \tfrac{c-e}{d-f}$ 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.
 A: 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
$$\frac{a}{b}=\frac{e}{f}=\frac{c-e}{d-f}.$$
A: 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. 
A: assuming that none of the denominators are zero :
$$\frac{a}{b} = \frac{c}{d} \rightarrow ad=bc$$
$$\frac{a}{b} = \frac{e}{f} \rightarrow af=be$$
$$ad-af=bc-be$$
$$a(d-f)=b(c-e)\rightarrow \frac{a}{b} = \frac{c-e}{d-f}$$
A: $\begin{array}{rrrl}{\bf Hint} 
&& d\, x\!\!\! &=&\!\!\! c\\
&& f\, x \!\!\!&=&\!\!\! e\\
\Rightarrow &&(d\!-\!f)\, x\!\!\! &=&\!\!\! c\!-\!e\ \ \text{ by subtracting the prior equations.}\end{array}$
Geometrically: if vectors $(d,c),\,(f,e)$ have the same slope, then so too does their difference.
