Most students asked why is that ${a\over b}\div{c\over d}={ad\over bc}$

By definition, $x \div y$ should be a number $z$ such that $x = y z$.
So you just verify: $$ \dfrac{ad}{bc} \times \dfrac{c}{d} = \dfrac{a}{b}$$ Ultimately this works because of the associative and commutative laws of multiplication, but you don't have to tell the students that.

"Why" should always be allowed.

Depending on how much these students know, I'm a fan of this explanation:

$$\frac ab \div \frac cd = \frac{\frac ab}{\frac cd} = \frac{\frac {ad}{b}}{\frac {cd}{d}} = \frac{\frac {ad}{b}}{c} = \frac{\frac{ad}{bc}}{\frac cc} = \frac{ad}{bc} = \frac ab \cdot \frac dc$$

The only thing they have to accept is that we can multiply any fraction by $\frac dd$ or $\frac {\frac{1}{c}}{\frac{1}{c}}$, which should be obvious to them if they're doing fraction multiplication.

The motivation is there, too: we first get rid of the pesky $d$ from the denominator by multiplying by it, and then we get rid of the pesky $c$ by dividing by it: all in all, we've multiplied our $\frac ab$ by $\frac dc$.

I think the merit to this method is that it moves away from the notion of the $\div$ symbol and expresses the result as a "fraction of fractions," which is much more generalizable.

When you divide $x÷y$, it is the same as $\frac{x}{y}=x×\frac{1}{y}$. Students familiar with fractions should be able to understand this. This is true no matter what $x$ or $y$ are, even if they're fractions. Therefore, $\frac{a}{b}÷\frac{c}{d}=\frac{a}{b}×\frac{d}{c}=\frac{ad}{bc}$.