Given cyclic quadrilateral ABCD with sides of length a, b, c, and d. Let the length of diagonal BD be e. Let the angle between sides a and d have measure A, and let the angle between sides b and c have measure C.

Since quadrilateral ABCD is cyclic, then opposite angles A and C are supplementary (A + C = 180 degrees). Therefore, using trigonometry, cosA = -cosC, and sinA = sinC.

Using the Law of Cosines on the two triangles formed after
constructing diagonal e, e^{2} = a^{2} + d^{2}
- 2adcosA and e^{2} = b^{2} + c^{2} -
2bccosC. By transitivity, a^{2} + d^{2} - 2adcosA
= b^{2} + c^{2} - 2bccosC. Since cosA = -cos C,
then a^{2} + d^{2} - 2adcosA = b^{2} +
c^{2} + 2bccosA. Therefore, a^{2} - b^{2}
- c^{2} + d^{2} = 2(ad + bc)cosA. Squaring each
side of the equation gives (a^{2} - b^{2} - c^{2}
+ d^{2})^{2} = 4(ad + bc)^{2}*(cosA)^{2}.

Using the trigonometric formula for finding the area of a triangle,
and since the area of the quadrilateral is the sum of the area
of the two triangles formed after constructing diagonal e, then
the area of the quadrilateral equals (1/ 2)adsinA + (1/ 2)bcsinC.
By substitution, Area = (1/ 2)adsinA + (1/ 2)bcsinA. Therefore,
Area = (1/ 2)(ad + bc)sinA, and so 2*Area = (ad + bc)sinA. Multiplying
both sides of the equation by 2 gives 4*Area = 2(ad + bc)sinA.
Squaring both sides of the equation gives 16(Area)^{2}
= 4(ad + bc)^{2}*(sinA)^{2}.

By the addition property of equality, (a^{2} - b^{2}
- c^{2} + d^{2})^{2} + 16(Area)^{2}
= 4(ad + bc)^{2}*(cosA)^{2} + 4(ad + bc)^{2}*(sinA)^{2}.
Factoring on the right side of the equation gives (a^{2}
- b^{2} - c^{2} + d^{2})^{2} +
16(Area)^{2} = 4(ad + bc)^{2}*[(cosA)^{2}
+ (sinA)^{2}]. Substituting the trigonometric identity
(sinA)^{2} + (cosA)^{2} = 1 gives (a^{2}
- b^{2} - c^{2} + d^{2})^{2} +
16(Area)^{2} = 4(ad + bc)^{2}. Therefore, 16(Area)^{2}
= 4(ad + bc)^{2} - (a^{2} - b^{2} - c^{2}
+ d^{2})^{2}. Note: For the rest of the proof,
let A represent the area of the quadrilateral.

Factoring the right side of the equation gives 16A^{2}
= [2(ad + bc) + (a^{2} - b^{2} - c^{2}
+ d^{2})]*[2(ad + bc) - (a^{2} - b^{2}
- c^{2} + d^{2})]. After a little regrouping,
16A^{2} = [(a^{2} + 2ad + d^{2}) - (b^{2}
- 2bc + c^{2})]*[(b^{2} + 2bc + c^{2})
- (a^{2} - 2ad + d^{2})]. After factoring again,
16A^{2} = [(a + d)^{2} - (b - c)^{2}]*[(b
+ c)^{2} - (a - d)^{2}]. Factoring once more gives
16A^{2} = [(a + d) + (b - c)]*[(a + d) - (b - c)]*[(b
+ c) + (a - d)]*[(b + c) - (a - d)]. Therefore, 16A^{2}
= (a + b - c + d)(a - b + c + d)(a + b + c - d)(-a + b + c + d).

Since the semiperimeter of the above cyclic quadrilateral is defined as s = (a + b + c + d)/ 2, it follows that 2s = a + b + c + d, and so 2s - a = b + c + d. Adding -a to both sides of the equation gives 2s - 2a = -a + b + c + d. Therefore, 2(s - a) = -a + b + c + d. Similar algebra steps give 2(s - b) = (a - b + c + d), 2(s - c) = (a + b - c + d), and 2(s - d) = (a + b + c - d).

By substitution, 16A^{2} = [2(s - c)]*[2(s - b)]*[2(s
- d)]*[2(s - a)]. After a little rearranging, 16A^{2}
= 16(s - a)(s - b)(s - c)(s - d). Finally, A^{2} = (s
- a)(s - b)(s - c)(s - d), and so A = [(s - a)(s - b)(s - c)(s
- d)]^{1/2}, which was to be proved. QED.