proof of Brahmagupta’s formula

We shall prove that the area of a cyclic quadrilateral with sides $p,q,r,s$ is given by

$$\sqrt{(T-p)(T-q)(T-r)(T-s)}$$

where $T=\frac{p+q+r+s}{2}.$

Area of the cyclic quadrilateral = Area of $\mathrm{△}ADB+$ Area of $\mathrm{△}BDC.$

$$=\frac{1}{2}pq\sin A+\frac{1}{2}rs\sin C$$

But since $ABCD$ is a cyclic quadrilateral, $\mathrm{\angle }DAB={180}^{\circ }-\mathrm{\angle }DCB.$Hence $\sin A=\sin C.$ Therefore area now is

$$Area=\frac{1}{2}pq\sin A+\frac{1}{2}rs\sin A$$

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

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

$$4{(Area)}^2={(pq+rs)}^2-co{s}^{2}A{(pq+rs)}^2$$

Applying cosines law for $\mathrm{△}ADB$ and $\mathrm{△}BDC$ 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 $(a+b)(a-b)$ as

$$(2(pq+rs)+p^2+q^2-r^2-s^2)(2(pq+rs)-p^2-q^2+r^2+s^2)$$

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

$$=(p+q+r-s)(p+q+s-r)(p+r+s-q)(q+r+s-p)$$

Introducing $T=\frac{p+q+r+s}{2},$

$$16{(Area)}^2=16(T-p)(T-q)(T-r)(T-s)$$

Taking square root, we get

$$Area=\sqrt{(T-p)(T-q)(T-r)(T-s)}$$