Prove $\dfrac{ax+b}{cx+d}=\dfrac{b}{d}$ if $ad=bc$. Well obviously if $\dfrac{ax+b}{cx+d}=\dfrac{b}{d}$ holds then $cbx+bd=bd+adx$ and it holds for any $x$ if $ad=bc$.
However, my question is to 'algebraically' or 'directly' calculate the $\dfrac{ax+b}{cx+d}$ if $ad=bc$. By multiplying both denominator any numerator by any possible guess it doesn't reduce to a constant! 
Please help! Thank you.

If $ad=bc$ then $\frac{a}{b}=\frac{c}{d}$.
Let's denote this value by $\beta$. Then we have $a=\beta b\text{ and }c=\alpha d\text.$
So then $$\frac{ax+b}{cx+d}=\frac{\beta bx + b}{\beta dx + d}=\frac{b(\beta x+1)}{d(\beta x+1)} = \frac{b}{d}\text.$$
