ব্রহ্মগুপ্তের সূত্র

Brahmagupta's Formula

Well, now that you've learned heron's (bashy) formula, there's another (more bashy) formula for the area of a cyclic quadrilateral given its side lengths! How nice.

Brahmagupta's Formula states:

The area of a cyclic quadrilateral $ABCD$ with sides of length $a, b, c, d$ and semiperimeter $s$ is equal to $$\sqrt{(s-a)(s-b)(s-c)(s-d)}$$

Proof

Denote the area of polygon $A_1A_2\dots A_n$ by $[A_1A_2\dots A_n]$. We let $a=AB$, $b=BC$, $c=CD$, and $d=AD$. Then, we have: $$[ABCD] = [ABD] + [CBD] = \dfrac 12 ad\sin A + \dfrac 12 bc\sin C$$ Since $A = 180^\circ - C$, we have $\sin A = \sin C$, so the above expression simplifies to: $$[ABCD] = \dfrac 12(ad + bc) \sin A$$ $$[ABCD]^2 = \dfrac 14(ad + bc)^2 \sin^2A$$ $$4[ABCD]^2 = (ad + bc)^2 (1 - \cos^2A)$$ $$4[ABCD]^2 = (ad + bc)^2 - (ad + bc)^2\cos^2A$$ We see a cosine and we want to get rid of it, so we apply the law of cosines on $\triangle ADB$ and $\triangle CDB$ to get $DB$ and set them equal to end up with: $$a^2 + d^2 - 2ad\cos A = b^2 + c^2 - 2bc\cos C$$ $$a^2 + d^2 - 2ad\cos A = b^2 + c^2 + 2bc\cos A$$ $$a^2 + d^2 - b^2 - c^2 = 2(ad+bc)\cos A$$ $$\dfrac{a^2 + d^2 - b^2 - c^2}2 = (ad+bc)\cos A$$ Then we can substitute this to the area formula to get: $$4[ABCD]^2 = (ad + bc)^2 - \dfrac 14(a^2 + d^2 - b^2 - c^2)^2$$ $$16[ABCD]^2 = (2ad + 2bc)^2 - (a^2 + d^2 - b^2 - c^2)^2$$ We then use difference of squares (and rearrange a bit): $$16[ABCD]^2 = (a^2 + 2ad + d^2 - (b^2 - 2bc + c^2)) \cdot \qquad \qquad \qquad (-(a^2 - 2ad + d^2) + b^2 + 2bc + c^2)$$ $$16[ABCD]^2 = ((a+d)^2 - (b-c)^2)((b+c)^2 - (a-d)^2)$$ Now, we can apply difference of squares again to get: $$\begin{align} 16[ABCD]^2 &= (a+d+b-c)(a+d-b+c) \cdot \\ &\quad (b+c+a-d)(b+c-a+d) \\ &= (-a+b+c+d)(a-b+c+d) \cdot \\ &\quad (a+b-c+d)(a+b+c-d) \end{align}$$ Now, this would be a nice formula already, but let's try to simplify it a bit to get those semiperimeters in. We plug in semiperimeters to get: $$16[ABCD]^2 = 16(s-a)(s-b)(s-c)(s-d)$$ Then, dividing by 16 and taking the square root, we get: $$[ABCD] = \sqrt{(s-a)(s-b)(s-c)(s-d)}$$ As desired.

Also, note that by taking limit as $D \rightarrow C$ we can get the herons formula back.