What does it mean that two pairs of segments are proportional? 
What does it mean that two pairs of segments are proportional?

Hi.
According to my book, a proportion of segments is defined like this:

Let $\overline{AB}, \overline{CD}, \overline{EF}, \overline{GH}$ be segments; then 
$\overline{AB}, \overline{CD}$ are proportional to $\overline{EF},
 \overline{GH}$ ⇔ $\frac{AB}{CD}=\frac{EF}{GH}$.

*The name of a line without an overline refers to the lenght of it.
Now, when talking about congruence, I know that $\overline{CD}≅\overline{EF}$ means, according to the axioms of the euclidean geometry course that I'm taking, which is not totally synthetic since it uses real numbers to characterize the length of segments, means that the lenght of $\overline{CD}$ and $\overline{EF}$ can be represented with the same real number, this is, they have the same lenght. However, I don't understand what does it mean that a pair of segments is congruent to another pair; Firstly I thought that 
$\frac{AB}{CD}=\frac{EF}{GH}$ meant that $AB=EF$ and $CD=GH$, 
but this wouldnt make sense, given that $\frac{4}{2}=\frac{2}{1}$.
So my questions is "What does it mean that two pairs of segments are proportional?"
Thanks in advance.
 A: It doesn't imply that AB=EF or that CD=HG. They are proportional because their quotients are equal. As you noted (AB/CD)=(EF/GH), such that ABGH=CDEF. In this case (4*1)=(2*2).
A: It's just like with numbers.
$\frac 37 = \frac 9{21}$ obviously does not mean $3 = 9$ but it means that there is as value, a proportional scale $a$, so that $9 = 3*a$ and $21= 7*a$ and $\frac 37 = \frac 9{21}= \frac {3a}{7a}$.  (In this case $a = 3$.)
Note, the proportional scale need not be an integer.  It'd be just as correct to say, the proportional scale $r$ so that $3 = 9*r$ and $7 = 21*r$ and $\frac 37 = \frac {9}{21} = \frac {9r}{21r}$.  (In that case $r = \frac 13$.)
So $\frac {AB}{CD} = \frac {EF}{GH}$ means that if $AB = s*CD$ for some $s$ then $EF = s*GH$ for the exact same value.
So if $AB = 6$ and $CD= 2$ and $EF = 15$ and $GH = 5$.  Then $AB$ is proportional to $CD$ (it is three times larger) by the same rate as $EF$ is proportional to $GH$ (it is three times longer)
Not that it also means that $AB$ is proportional to $EF$ (it is $\frac 25$ as long) as $CD$ is to $GH$ (it is $\frac 25$ as long).
"$AC$, $CD$ are proportional to $EF$, $GH$ $\iff \frac {AC}{CD} = \frac{EF}{GH}\iff \frac {AC}{EF} = \frac {CD}{GH} $.  Further more if $\frac {AC}{CD} = \frac {EF}{GH} = a$ and $\frac {AC}{EF} = \frac {CD}{GH} = r$ then $AC = r*EF$ and $CD = r*GH$ while ${AC} = a*CD$ and $EF = a*GH$.
