Let $ABCD$ be a cyclic quadrilateral with side lengths $AB=p,BC=q,CD=r,DA=s$.Show that $\dfrac{AC}{BD}=\dfrac{ps+qr}{pq+rs}$. 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.
 A: As you've said this follows immidiatelly from the Second Ptolemy Therem. Here's the proof for it.
Note that in every triangle we have:
$$bc = 2Rh_{a}$$
This follows from the formula for area of triangle: $P = \frac{abc}{4R}$

Note: Vertex C and vertex D need to switch places.
So from $\triangle ABD$ and $\triangle BCD$ in the picture we have:
$$ad = 2Rh_1 \quad \quad \text { and } \quad \quad {bc = 2Rh_2}$$
Adding this we have:
$$ad + bc = 2Rh_1 + 2Rh_2$$
From the right trinagles, where $AK$ and $KD$ are hypothenyses we have:
$$h_1 = AK \cdot \sin w \quad \quad \text { and } \quad \quad {h_2 = KC \cdot \sin w}$$
Substitunting we have:
$$ad + bc = 2R \sin w (AK + KC) = 2R \cdot AC \sin w$$
Simularly we have:
$$ab + cd = 2R \cdot BD \sin (\pi - w)$$
But $\sin w = \sin (\pi - w)$, so we have:
$$\frac{ad + bc}{ab + cd} = \frac{2R \cdot AC \sin w}{2R \cdot BD \sin (\pi - w)} = \frac{AC}{BD}$$
A: Recall that in a cyclic quadrilateral $\square ABCD$, opposite angles are supplementary (and, therefore, have a common sine). Also, as a consequence of the Law of Sines, a triangle $\triangle XYZ$ inscribed in a circle of diameter $d$ has side-lengths
$$|YZ| = d \sin X \qquad |ZX| = d \sin Y \qquad |XY| = d \sin Z$$
For $\square ABCD$ inscribed in a circle of diameter $d$, then, we can write
$$|BD| = d \sin A = d \sin C \qquad |AC| = d \sin B = d\sin D$$
Now,
$$\begin{align}
ps+ qr &= \frac{2}{|BD|/d} \left(\; \frac{1}{2} p s \sin A + \frac{1}{2} qr \sin C \;\right) \\[4pt]
&= \frac{2d}{|BD|} \left( |\triangle BAD| + |\triangle BCD| \right) \\[4pt]
&= \frac{2d |\square ABCD|}{|BD|} \\[6pt]
pq+rs &= \frac{2d|\square ABCD|}{|AC|}
\end{align}$$
whence
$$\frac{ps+qr}{pq+rs} = \frac{|AC|}{|BD|}$$
A: Area of a triangle of sides $ {a, b, c} $ inscribed in a circle of radius $ {R} $ is $ {{{ K}}= {\frac {abc} {4R}}} $
Now, let us consider a quadrilateral with sides ${ AB=p , BC=q , CD=r , DA=s }$.

In Figure 1 :
       Divide the quadrilateral into two triangles with area K1 and K2 respectively.Let the area of the quadrilateral be ${Q}$.       
Using the formula for Area of Triangle,
    $ {K1=\frac {BD.p.s}{4R}} $ 
${K2=\frac{BD.q.r}{4R}}$ 
So, ${ Q= K1+K2=\LARGE{{\frac {BD.(ps+qr)} {4R}}}}$     ....(1)
Similarly in Figure 2 :
${K3=\frac {AC.r.s} {4R}}$
${K4=\frac {AC.p.q} {4R}}$
So, ${ Q= K3+K4=\LARGE{{\frac {AC.(rs+pq)} {4R}}}}$     ....(2)
Now, from equation ${(1)}$ & ${(2)}$,
${{AC.(pq+rs)=BD.(ps+qr)}}$
Hence proved,
${\LARGE{{\frac {AC}{BD}} = {\frac {ps + qr}{pq + rs}}}}$
