${\bf METHOD}$ ${\bf 1}$
First she applied Brahmagupta's formula for the area of a
quadrilateral:
$ \delta = \sqrt{(s  a)(s  b)(s
 c)(s  d)  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 $abcd \cos^2
\beta$ is least. Since $\cos^2 \beta$ is always positive, this
value is least when $\beta $ is $90$ degrees as $\cos 90^{\circ} =
0$. Hence $\frac{1}{2}(A + C) = 90 $ and so $A + C = 180$ showing
that the opposite angles in the quadrilateral add up to
$180^{\circ}$ 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.
${\bf METHOD}$ ${\bf 2}$ The area of the quadrilateral $ABCD$ can be expressed as the
sum of the areas of triangle $ABD$ and $BCD$. Let the area of
$ABCD$ be $\Delta$. Then \[ \Delta = \frac{1}{2} ad\sin A +
\frac{1}{2} bc \sin C \] Thus $$\frac {d\Delta}{dA} = \frac{1}{2}
ad\cos A + \frac{1}{2} bc \cos C \frac {dC}{dA}.\quad (1)$$
From the Cosine Rule, $$a^2 + d^2  2ad\cos A = b^2 + c^2 
2bc\cos C,$$
Hence, (differentiating both sides with respect to $A$),
$$2ad\sin A = 2bc\sin C \frac {dC}{dA}.\quad (2)$$
From (1) and (2), $$\eqalign{ \frac{d\Delta}{dA} = \frac{1}{2}
ad\cos A + \frac{1}{2} bc \cos C{ad\sin A \over bc\sin C}\\ = \frac
{ad\cos A \sin C + \sin A \cos C}{2\sin C}\\ =
\left(\frac{ad}{2}\right) \frac{\sin (A + C)}{\sin C} }.$$ 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.
