ব্রেৎশ্নাইডারের সূত্র

Bretschneider's Formula

Okay, so far we've seen two bashy formulas for areas of general triangles and cyclic quadrilaterals. Now, we still don't know the area of a general quadrilateral. Here, we will prove a formula for the area of a general quadrilateral given its side lengths (this is extended from brahmagupta's formula similarly to how the law of cosines is extended from the pythagorean theorem).

So, without more words to delay, here's the formula:

The area of general quadrilateral $ABCD$ with sides of length $a,b,c,d$, semiperimeter $s$, and two opposite angles $\alpha$ and $\beta$ is $$\sqrt{(s-a)(s-b)(s-c)(s-d) - abcd\cos^2\left(\dfrac{\alpha + \beta}2\right)}$$ Wow, really mean looking. I almost didn't even have enough horizontal space to put it in this explanation! However, as we'll see, this follows from essentially the same argument as brahmagupta's formula did.

Proof

Denote the area of polygon $A_1A_2\dots A_n$ by $[A_1A_2\dots A_n]$. We let $\angle A = \alpha$, $\angle C = \beta$, $a=AB$, $b=BC$, $c=CD$, and $d=AD$. We note that $[ABCD] = [ABD] + [CBD]$, so: $$[ABCD] = \dfrac 12(ad \sin\alpha + bc \sin\beta)$$ Then, $$4[ABCD]^2 = (ad)^2 \sin^2\alpha + (bc)^2 \sin^2\beta + 2abcd\sin\alpha\sin\beta$$ By the law of cosines on $\triangle ABD$ and $\triangle CBD$, we have: $$a^2 + d^2 - 2ad\cos\alpha = b^2 + c^2 - 2bc\cos\beta$$ We then put all the cosines on one side, divide by 2, and square: $$\dfrac{(a^2 + d^2 - b^2 - c^2)^2}4 = (ad)^2\cos^2\alpha + (bc)^2\cos^2\beta - 2abcd\cos\alpha\cos\beta$$ We then add this to the area formula and use the identity $\sin^2\theta + \cos^2\theta = 1$ and the cosine angle sum identity: $$4[ABCD]^2 + \dfrac{(a^2 + d^2 - b^2 - c^2)^2}4 = \\ (ad)^2 + (bc)^2 - 2abcd\cos(\alpha + \beta)$$ Then: $$4[ABCD]^2 + \dfrac{(a^2 + d^2 - b^2 - c^2)^2}4 = \\ (ad + bc)^2 - 4abcd\cos^2\left(\dfrac{\alpha + \beta}2\right)$$ Following the same steps as in our proof of brahmagupta's formula, we get: $$16[ABCD]^2 = 16(s-a)(s-b)(s-c)(s-d) - 16abcd\cos^2\left(\dfrac{\alpha+\beta}2\right)$$ Then, we can divide and take the square root of both sides to get that $[ABCD]$ equals: $$\sqrt{(s-a)(s-b)(s-c)(s-d) - abcd\cos^2\left(\dfrac{\alpha + \beta}2\right)}$$ So we are done. Whew!