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.

Multiplying 
$$\frac{ax+b}{cx+d}$$
by $\frac{bd}{bd}\ (=1)$ gives
$$\frac{b(adx+bd)}{d(bcx+bd)}=\frac{b(bcx+bd)}{d(bcx+bd)}\tag1$$
since $ad=bc$.
Now $(1)$ equals $\frac bd$.
