任意四边形的面积公式

定理

在四边形 $ABCD$ 中，记 $|AB| = a, |BC| = b, |CD| = c, |DA| = d$, 再记半周长 $p = \dfrac{a + b + c + d}{2}$. 则四边形 $ABCD$ 的面积为 $$ \sqrt{(p - a)(p - b)(p - c)(p - d) - abcd \cos^2 \dfrac{A + C}{2}} $$

证明. 连接 $B, D$, 则 $$ Area = \dfrac{1}{2} ad \sin A + \dfrac{1}{2} bc \sin C $$

因此 $$ \begin{align} 4Area^2 & = a^2d^2 \sin^2 A + b^2c^2 \sin^2 C + 2abcd \sin A \sin C \notag \ & = a^2d^2 + b^2c^2 - a^2d^2 \cos^2 A - b^2c^2 \cos^2 C + 2abcd \sin A \sin C \end{align} $$

在 $\triangle ABD, \triangle BCD$ 中用余弦定理得 $$ a^2 + d^2 - 2ad \cos A = |BD|^2 = b^2 + c^2 - 2bc \cos C $$

因此 $$ a^2d^2 \cos^2 A + b^2c^2 \cos^2 C = \dfrac{1}{4} (a^2 + d^2 - b^2 - c^2)^2 + 2abcd \cos A \cos C $$

代入得 $$ \begin{align*} 16Area^2 & = 4a^2d^2 + 4b^2c^2 - \left( a^2 + d^2 - b^2 - c^2 \right)^2 - 8abcd(\cos A \cos C - \sin A \sin C) \ & = 4a^2d^2 + 4b^2c^2 - \left( a^2 + d^2 - b^2 - c^2 \right)^2 - 8abcd \cos (A + C) \ & = 4a^2d^2 + 4b^2c^2 + 8abcd - \left( a^2 + d^2 - b^2 - c^2 \right)^2 - 16abcd \cos^2 \dfrac{A + C}{2} \ & = \left( 2ad + 2bc \right)^2 - \left( a^2 + d^2 - b^2 - c^2 \right)^2 - 16abcd \cos^2 \dfrac{A + C}{2} \ & = \left( 2ad + 2bc + a^2 + d^2 - b^2 - c^2 \right) \left( 2ad + 2bc - a^2 - d^2 + b^2 + c^2 \right) - 16abcd \cos^2 \dfrac{A + C}{2} \ & = \left( (a + d)^2 - (b - c)^2 \right) \left( (b + c)^2 - (a - d)^2 \right) - 16abcd \cos^2 \dfrac{A + C}{2} \ & = (a + b + c - d)(b + c + d - a)(c + d + a - b)(d + a + b - c) - 16abcd \cos^2 \dfrac{A + C}{2} \ & = 16(p - a)(p - b)(p - c)(p - d) - 16abcd \cos^2 \dfrac{A + C}{2} \ & = 16 \left( (p - a)(p - b)(p - c)(p - d) - abcd \cos^2 \dfrac{A + C}{2} \right) \end{align*} $$

故 $Area = \sqrt{(p - a)(p - b)(p - c)(p - d) - abcd \cos^2 \dfrac{A + C}{2}}$.

注记

1. 由 $A + B + C + D = 2 \pi$ 知 $\cos^2 \dfrac{A + C}{2} = \cos^2 \dfrac{B + D}{2}$, 因此还有 $$ Area = \sqrt{(p - a)(p - b)(p - c)(p - d) - abcd \cos^2 \dfrac{B + D}{2}} $$
2. $Area$ 的最大值为 $\sqrt{(p - a)(p - b)(p - c)(p - d)}$, 当且仅当 $A, B, C, D$ 四点共圆时取等.