show that if $\frac{a}{b} = \frac{c}{d}$ then $\frac{a}{b}=\frac{c}{d} = \frac{a-c}{b-d}$

Question

$\frac{a}{b}=\frac{c}{d} = \frac{a-c}{b-d}$

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)

Answer 1

$\dfrac{a}{b} = \dfrac{c}{d} \Rightarrow ad = bc \Rightarrow ad - cd = bc - cd \Rightarrow (a-c)d = (b-d)c \Rightarrow \dfrac{a-c}{b-d} = \dfrac{c}{d} = \dfrac{a}{b}$

Answer 2

Hint: Asserting that $\frac ab=\frac cd$ is equivalent to $ad=bc$. Now, use this equality to prove that $\frac{a-c}{b-d}=\frac ab$.

Answer 3

Substituting $$a=bt,c=dt$$ then $$\frac{a-c}{b-d}=\frac{bt-dt}{b-d}=t$$

Answer 4

$\dfrac a b - \dfrac {a-c}{b-d} = \dfrac {a(b-d)-b(a-c)}{b(b-d)}=\dfrac{bc-ad}{b(b-d)}=\dfrac{c-a\dfrac d b }{b-d}=\dfrac{c-a\dfrac c a}{b-d}=0$

Answer 5

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).

Answer 6

$$\begin{align} \text{If }\qquad \frac ab=\frac cd&=m\\ \text{then}\qquad\qquad a&=mb\\ c&=md\\ \therefore \frac {\lambda a+\mu c}{\lambda b+\mu d} &=\frac{\lambda mb+\mu md}{\lambda b+\mu d}\\ &=\frac {m \big(\lambda b+\mu d\big)}{\lambda b+\mu d}\\ &=m\\ \end{align}$$