$\frac{a}{b} = \frac{c}{d} $ if $ad = cb$, how to intuitively understand this? This works if you multiply both sides with $bd$ and cancel stuff out... But how does it work? 
When I look at it, I would never guess something like that is valid without resorting to the established arithmetic rules.
Maybe this is a nonsense question, should these things be analyzed in such a way or just accepted from the arithmetic rules? Since, that's the reason humanity has developed mathematics. To simplify and abstract things which would otherwise be out of the reach of our mind. 
Just a simple example, so, please... Set me straight. Thanks! 
 A: If $\frac{a}{b}$ represents "the solution to the equation $bx=a$", then saying that $\frac{c}{d}=\frac{a}{b}$ means that any solution to $bx=a$ is a solution to $dy=c$, and vice-versa. So if $x$ is a solution to $bx=a$, then multiplying by $d$ we have $ad = dbx = b(dx)$. But since $x$ is also a solution to $dy=c$, that means that $dx=c$, so $ad=b(dx) = bc$. 
So if $\frac{a}{b}=\frac{c}{d}$, then $ad=bc$.
Conversely, if $ad=bc$, and $x$ is a solution to $bx=a$, then it is also a solution to $dbx = da=bc$. Since $b\neq 0$, $dbx = bc$ if and only if $dx=c$, so $x$ is a solution to $bx=a$ if and only if it is a solution to $cy=d$.
In short, the equations $bx=a$ and $cy=d$, with $a,b,c,d$ integers, $b$ and $d$ nonzero, have the same solution if and only if $ad=bc$. So if $\frac{r}{s}$ for integers $r,s$, $s\neq 0$, represents "the solution to $sx=r$", then for integers $a,b,c,d$, $b\neq 0$, $d\neq 0$, 
$$\frac{a}{b}=\frac{c}{d}\text{ if and only if }ad=bc.$$
A: Would a picture help? 
Note that $\frac{a}{b} = \frac{c}{d}$ by similar triangles.  The blue and green rectangle has area $ad$ while the green and yellow rectangle has area $bc$.  These are equal, namely fraction $\frac{a}{b} = \frac{c}{d}$ 
of the area of the big rectangle.
A: HINT $\ $ It boils down to putting the two fractions over the common denominator $\rm\:b\:d\:,\:$ or, equivalently, changing the "unit" of measurement on your ruler from $1$ to $\rm\:1/(b\:d)\:.$   
On the new ruler $\rm\ \dfrac{1}b\ $ has measure $\rm\ d\ $ since $\rm\ \dfrac{1}b\: =\ d\:\dfrac{1}{b\:d}\ $ hence $\rm\ a\:\dfrac{1}b\ $ has measure $\rm\ a\:d\:.$  
Similarly $\rm\ c\dfrac{1}d\ $ has measure $\rm\:c\:b\:.$
Analogously, you can use this ruler to compare any fractions whose denominator divides $\rm\:b\:d\:.$
A: <------a------> <--d-->
<---b---> <-----c----->
         ^     ^
         | a-b |
         | c-d |  

That is a+d=b+c <-> a-b=c-d. Then, assume this was drawn on logarithmic ruler; therefore, substitute "+" for "*" and "-" for "/"
A: Lately, I've been thinking about it this way:
If $\frac{a}{b} = \frac{c}{d} $ then $ \frac{c}{d} $ must be equal to $\frac{ka}{kb} $. Their ratio is the same if and only if $a$ and $b$ are scaled by a constant $k$. 
Therefore $\frac{a}{b} = \frac{c}{d} $ can be rewritten as $\frac{a}{b} = \frac{ka}{kb} $.
If the constant $k$ is the same, which is the condition of equality, we can simply cancel it out, which will leave us with $\frac{a}{b} = \frac{a}{b} $ which is evidently true. 
If we didn't want to cancel out the constant $k$, we can try the $ad = bc$ method. Let's move $kb$ and $b$ to the other sides of the equation:
$akb = kab$ -> $akb = akb$ 
which says that if the constant in the numerator and the denominator is different, the equality fails. Is this also a way of looking at this?
A: An intuitive way is to think of a money problem. $a$ is the money that John produced in one hour then $ad$ is the total after $d$ hours. Then let $c$ be the money that Paul produced. 
Paul wants to work enough hours $d$ to produce the same rate of money like John. Then Paul needs to work $\frac{a}{c}d$ hours to produce the same amount per hour like John.
A: You may wish to read through the comments/answers to this related SE question:
Is there any toy for learning algebraic manipulation of fractions?
In particular, I think it helps to think of transforming fractions using the single rule:  If a quantity moves across the $=$ sign, it must change positions in the fraction -- numerator becomes denominator, and vice versa.
Hope this helps!
