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.

Note that $d(ax+b)=(ad)x+bd=(bc)x+bd=b(cx+d)$. From here, the conclusion is trivial.
