Maximum area of a quadrilateral

Using Bretschneider's formula ,

$\sqrt{(s-a)(s-b)(s-c)(s-d)-abcd \cos^{2} \left( \frac{ \alpha+\gamma}{2} \right) }$

Here $\alpha$ and $\gamma$ are two opposite angles of the quadrilateral.

For the area to be maximum,

$abcd \cos^{2} \left( \frac{ \alpha+\gamma}{2} \right)$

This expression must be minimum and only in a cyclic quadrilateral opposite angle's sum is $180^\circ$ . Therefore,

$abcd \cos^{2} \left( \frac{ \alpha+\gamma}{2} \right)=0$

And the area is thus maximum for cyclic quadrilateral.

I thought this must be provided to make solution complete.