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

d = Ö   (s - a)(s - b)(s - c)(s - d) - abcd cos2 b

where

s =

1 2

(a + b + c + d)

and

b =

1 2

(A + C)

or

1 2

(B + D)

.

Hence the area is clearly the greatest when abcd cos² b is least. Since cos² b is always positive, this value is least when b is 90 degrees as cos90^o = 0. Hence

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 ABCD can be expressed as the sum of the areas of triangle ABD and BCD. Let the area of ABCD be D then

D = 1 2 adsinA + 1 2 bc sinC

Thus

dD dA = 1 2 adcosA + 1 2 bc cosC dC dA .    (1)

From the Cosine Rule,

a2 + d2 - 2adcosA = b2 + c2 - 2bccosC,

Hence, (differentiating both sideswith respect to A),

2adsinA = 2bcsinC dC dA .    (2)

From (1) and (2),

frac dDdA = 1 2 adcosA + 1 2 bc cosC adsinA bcsinC = adcosA sinC + sinA cosC 2sinC = æ ç è ad 2 ö ÷ ø sin(A + C) sinC .

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

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