show that if $\frac{a}{b} = \frac{c}{d}$ then $\frac{a}{b}=\frac{c}{d} = \frac{a-c}{b-d}$ $\frac{a}{b}=\frac{c}{d} = \frac{a-c}{b-d}$
Precalculus Mathematics in a nutshell says it's easy to verify by cross-multiplication but I'm having difficulty doing so. 
(not a homework problem, any help would be appreciated) 
 A: Substituting $$a=bt,c=dt$$ then $$\frac{a-c}{b-d}=\frac{bt-dt}{b-d}=t$$
A: $\dfrac{a}{b} = \dfrac{c}{d} \Rightarrow ad = bc \Rightarrow ad -  cd = bc - cd \Rightarrow (a-c)d = (b-d)c \Rightarrow \dfrac{a-c}{b-d} = \dfrac{c}{d} = \dfrac{a}{b}$
A: Hint: Asserting that $\frac ab=\frac cd$ is equivalent to $ad=bc$. Now, use this equality to prove that $\frac{a-c}{b-d}=\frac ab$.
A: $\dfrac a b - \dfrac {a-c}{b-d} = \dfrac {a(b-d)-b(a-c)}{b(b-d)}=\dfrac{bc-ad}{b(b-d)}=\dfrac{c-a\dfrac d b }{b-d}=\dfrac{c-a\dfrac c a}{b-d}=0$
A: Simply subtract the fraction's (defining) equations, i.e.
$$\begin{align}   b\,y &\,=\, a\\ d\,y &\,=\, c_{\phantom{|}}\\ \hline \!\!\!\!\!\!\!\!\!\!\!\!\Rightarrow\ \ (b\!-\!d)\,y &\,=\, a\!-\!c\end{align}\qquad$$
Said geometrically: subtraction preserves same-slope vectors (here points on the line $\,y \,=\, (a/b)\,x)$
Remark $ $ This leads to a vector form of the Euclidean algorithm, and the descent step in a proof of Euclid's Lemma and the Fundamental Theorem of Arithmetic (essentially dating back to Euclid).  
A: $$\begin{align}
\text{If }\qquad  \frac ab=\frac cd&=m\\
\text{then}\qquad\qquad a&=mb\\
c&=md\\
\therefore \frac {\lambda a+\mu c}{\lambda b+\mu d}
&=\frac{\lambda mb+\mu md}{\lambda b+\mu d}\\
&=\frac {m \big(\lambda b+\mu d\big)}{\lambda b+\mu d}\\
&=m\\
\end{align}$$
