We shall prove that the area of a cyclic quadrilateral with sides $p,q,r,s$ is given by
 $\sqrt{(Tp)(Tq)(Tr)(Ts)}$ 

where $T=\frac{p+q+r+s}{2}.$
Area of the cyclic quadrilateral = Area of $\triangle ADB+$ Area of $\triangle BDC.$
 $=\frac{1}{2}pq\sin A+\frac{1}{2}rs\sin C$ 

But since $ABCD$ is a cyclic quadrilateral, $\angle DAB=180^{\circ}\angle DCB.$
Hence $\sin A=\sin C.$ Therefore area now is
 $Area=\frac{1}{2}pq\sin A+\frac{1}{2}rs\sin A$ 

 $(Area)^{2}=\frac{1}{4}\sin^{2}A(pq+rs)^{2}$ 

 $4(Area)^{2}=(1\cos^{2}A)(pq+rs)^{2}$ 

 $4(Area)^{2}=(pq+rs)^{2}cos^{2}A(pq+rs)^{2}$ 

Applying cosines law for $\triangle ADB$ and $\triangle BDC$ and equating the expressions for side $DB,$ we have
 $p^{2}+q^{2}2pq\cos A=r^{2}+s^{2}2rs\cos C$ 

Substituting $\cos C=\cos A$ (since angles $A$ and $C$ are suppplementary) and rearranging, we have
 $2\cos A(pq+rs)=p^{2}+q^{2}r^{2}s^{2}$ 

substituting this in the equation for area,
 $4(Area)^{2}=(pq+rs)^{2}\frac{1}{4}(p^{2}+q^{2}r^{2}s^{2})^{2}$ 

 $16(Area)^{2}=4(pq+rs)^{2}(p^{2}+q^{2}r^{2}s^{2})^{2}$ 

which is of the form $a^{2}b^{2}$ and hence can be written in the form $(a+b)(ab)$ as
 $(2(pq+rs)+p^{2}+q^{2}r^{2}s^{2})(2(pq+rs)p^{2}q^{2}+r^{2}+s^{2})$ 

 $=((p+q)^{2}(rs)^{2})((r+s)^{2}(pq)^{2})$ 

 $=(p+q+rs)(p+q+sr)(p+r+sq)(q+r+sp)$ 

Introducing $T=\frac{p+q+r+s}{2},$
 $16(Area)^{2}=16(Tp)(Tq)(Tr)(Ts)$ 

Taking square root, we get
 $Area=\sqrt{(Tp)(Tq)(Tr)(Ts)}$ 
