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Malhar Rapal While studying for a math competition, I encountered Brahmagupta’s formula, which states that the area of any cyclic quadrilateral (all vertices of the quadrilateral lie on a single circle) is equal to: where a, b, c, and d are the 4 side lengths of the quadrilateral and s is the quadrilateral’s semi-perimeter: (a + b + c + d)/2en.wikipedia.org/wiki/Heron%27s_formula. I was astounded by its similarity to Heron's Formula for triangles as well as Bretschneider's formula and wondered why Brahmagupta’s formula stood true for all cyclic quadrilaterals. In this article, I will derive Brahmagupta’s famous formula. Note: All angles in this article will be referred to in degrees. The diagram above depicts a cyclic quadrilateral with sides a, b, c, d and angles α, β, γ, and 𝛿. One property of cyclic quadrilaterals is that all opposite angles are supplementary. This means that α + β = 180, and γ + 𝛿 = 180. Using α + β = 180, we see that α = 180 - β and thus by apply cosine on both sides: cos(α)=cos(180-β). The difference formula for cosine states that cos(a-b) = cos(a)cos(b) + sin(a)sin(b) for any a and b. We can use this formula on cos(180-β)to get: cos(α) = cos(180)cos(β) + sin(180)sin(β). cos(180)= -1 and sin(180)=0 hence cos(α) = -cos(β). We also apply sine to both sides to get sin(α)=sin(180-β). The difference formula for sine is sin(a-b) = sin(a)cos(b) - sin(b)cos(a) hence sin(α)=sin(180)cos(β)-cos(180)sin(β). sin(180) = 0 and cos(180) = -1, hence sin(α)=sin(β). Now, we can split the cyclic quadrilateral into two triangles as shown above with a diagonal of length L. We can use cosine rule for triangles which states that a^2=b^2+c^2 - 2bc cos(A) for triangles with side lengths a, b, c and an angle A opposite to side a. One triangle is bounded by sides a, b, and L and the other triangle is bounded by c, d, L. By applying the cosine rule on both of these triangles, we can see that L^2=a^2+b^2-2abcos(α)and L^2=c^2+d^2-2cdcos(β). We can therefore state that for this cyclic quadrilateral, a^2+b^2-2abcos(α) = c^2+d^2-2cdcos(β) Since cos(α)=-cos(β), we can say that cos(β) = -cos(α). By replacing the occurrence of cos(β) with -cos(α) in the equation a^2+b^2-2abcos(α) = c^2+d^2-2cdcos(β), we get: a^2+b^2-2abcos(α) = c^2+d^2+2cdcos(α) . And by simple rearrangement and factoring: a^2+b^2-c^2-d^2=2cos(α)(ab+cd). Now, we can use the area of the two triangles to get the area of the entire cyclic quadrilateral. Since Area = 1/2 ab sin(C)for a triangle with sides a, b and angle C opposite to the third side (side that’s not a or b), we can see that the area of the triangle bounded by a, b, and L is 1/2 ab sin(α) and the area of the triangle bounded by c, d, and L is 1/2 cd sin(β) . Since the entire cyclic quadrilateral is composed from these two triangles, the area of the quadrilateral is the sum of the areas of the two triangles. Thus, we state that A = 1/2 ab sin(α) + 1/2 cd sin(β). Using sin(α)=sin(β), we replace sin(β) with sin(α)and rearrange to get A=(ab+cd)/2 sin(α) . And by making sin(α) the subject, we get sin(α)=2A/(ab+cd). Brahmagupta faced a problem. How could he relate sin(α)=2A/(ab+cd) with a^2+b^2-c^2-d^2=2cos(α)(ab+cd)? Since for any right angled triangle the sine of an angle is the opposite/hypotenuse and since sin(α)=2A/(ab+cd), the proof creatively formulates a right angled triangle as shown above with hypotenuse ab+cd, a side 2A opposite to an angle α. By using the Pythagorean theorem, the third side becomes: With any right angled triangle, cos(a)= adjacent/hypotenuse. Thus:
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