## Dr. McCrory, Instructor

### by Shannon Umberger

#### Note: This proof was adapted from a proof on page 58 in Geometry Revisited by H.S.M. Coxeter and S.L. Greitzer.

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, e2 = a2 + d2 - 2adcosA and e2 = b2 + c2 - 2bccosC. By transitivity, a2 + d2 - 2adcosA = b2 + c2 - 2bccosC. Since cosA = -cos C, then a2 + d2 - 2adcosA = b2 + c2 + 2bccosA. Therefore, a2 - b2 - c2 + d2 = 2(ad + bc)cosA. Squaring each side of the equation gives (a2 - b2 - c2 + d2)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, (a2 - b2 - c2 + d2)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 (a2 - b2 - c2 + d2)2 + 16(Area)2 = 4(ad + bc)2*[(cosA)2 + (sinA)2]. Substituting the trigonometric identity (sinA)2 + (cosA)2 = 1 gives (a2 - b2 - c2 + d2)2 + 16(Area)2 = 4(ad + bc)2. Therefore, 16(Area)2 = 4(ad + bc)2 - (a2 - b2 - c2 + d2)2. Note: For the rest of the proof, let A represent the area of the quadrilateral.

Factoring the right side of the equation gives 16A2 = [2(ad + bc) + (a2 - b2 - c2 + d2)]*[2(ad + bc) - (a2 - b2 - c2 + d2)]. After a little regrouping, 16A2 = [(a2 + 2ad + d2) - (b2 - 2bc + c2)]*[(b2 + 2bc + c2) - (a2 - 2ad + d2)]. After factoring again, 16A2 = [(a + d)2 - (b - c)2]*[(b + c)2 - (a - d)2]. Factoring once more gives 16A2 = [(a + d) + (b - c)]*[(a + d) - (b - c)]*[(b + c) + (a - d)]*[(b + c) - (a - d)]. Therefore, 16A2 = (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, 16A2 = [2(s - c)]*[2(s - b)]*[2(s - d)]*[2(s - a)]. After a little rearranging, 16A2 = 16(s - a)(s - b)(s - c)(s - d). Finally, A2 = (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.