METHOD 1
First she applied Brahmagupta's formula for the area of a quadrilateral.

${\displaystyle \delta =\sqrt{(s-a)(s-b)(s-c)(s-d)-\mathrm{abcd}{\cos }^2\beta }}$

where $s=\frac{1}{2}(a+b+c+d)$ and $\beta =\frac{1}{2}(A+C)$ or $\frac{1}{2}(B+D)$.

Hence the area is clearly the greatest when $\mathrm{abcd}{\cos }^2\beta$ is least. Since ${\cos }^2\beta$ is always positive, this value is least when $\beta$ is 90 degrees as $\cos {90}^o=0$. Hence $\frac{1}{2}(A+C)=90$ and so $A+C=180$ showing that the opposite angles in the quadrilateral add up to ${180}^o$ and so the area of a quadrilateral with fixed lengths of sides is greatest when it is cyclic.

This method gives a proof of the required result but you have to assume Brahmagupta's formula and Sue's second method uses only the formula for the area of a triangle.

METHOD 2
The area of the quadrilateral $\mathrm{ABCD}$ can be expressed as the sum of the areas of triangle $\mathrm{ABD}$ and $\mathrm{BCD}$. Let the area of $\mathrm{ABCD}$ be $\Delta$ then

${\displaystyle \Delta =\frac{1}{2}\mathrm{ad}\sin A+\frac{1}{2}\mathrm{bc}\sin C}$

Thus

${\displaystyle \frac{d\Delta }{\mathrm{dA}}=\frac{1}{2}\mathrm{ad}\cos A+\frac{1}{2}\mathrm{bc}\cos C\frac{\mathrm{dC}}{\mathrm{dA}}.\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}(1)}$

From the Cosine Rule,

${\displaystyle a^2+d^2-2\mathrm{ad}\cos A=b^2+c^2-2\mathrm{bc}\cos C,}$

Hence, (differentiating both sideswith respect to $A$),

${\displaystyle 2\mathrm{ad}\sin A=2\mathrm{bc}\sin C\frac{\mathrm{dC}}{\mathrm{dA}}.\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}(2)}$

From (1) and (2),

${\displaystyle \begin{array}{ccc}\multicolumn{1}{c}{\mathrm{frac}d\Delta \mathrm{dA}}& =& \frac{1}{2}\mathrm{ad}\cos A+\frac{1}{2}\mathrm{bc}\cos C\frac{\mathrm{ad}\sin A}{\mathrm{bc}\sin C}\\ \multicolumn{1}{c}{}& =& \frac{\mathrm{ad}\cos A\sin C+\sin A\cos C}{2\sin C}\\ \multicolumn{1}{c}{}& =& \left(\frac{\mathrm{ad}}{2}\right)\frac{\sin (A+C)}{\sin C}\end{array}.}$

Hence, for the maximum area, $\sin (A+C)=0$ and $A+C=\pi$ which makes the quadrilateral cyclic.

We can show that this gives the maximum and not the minimum value of $\Delta$ by finding the second derivative.