Let $ABCD$ be a cyclic quadrilateral with side lengths $AB=p,BC=q,CD=r,DA=s$.Show that $\dfrac{AC}{BD}=\dfrac{ps+qr}{pq+rs}$. 

My work:  

I have found out that this follows from Ptolemy's Second Theorem but cannot prove it.Please help! With or without Ptolemy is fine, I do not have any restriction.