海伦公式

设角 $C$ 夹在边 $a$ 与 $b$ 之间，对应对边为 $c$。则面积

$$A=\frac12 ab\sin C.$$

由余弦定理

$$\cos C=\frac{a^2+b^2-c^2}{2ab}.$$

因此

$$\sin^2 C=1-\cos^2 C=1-\left(\frac{a^2+b^2-c^2}{2ab}\right)^2.$$

代回得

$$A^2=\frac14 a^2 b^2\left[1-\left(\frac{a^2+b^2-c^2}{2ab}\right)^2\right] =\frac{1}{16}\left(4a^2b^2-(a^2+b^2-c^2)^2\right).$$

把右侧展开并整理，可因式分解为

$$A^2=\frac{1}{16}(a+b+c)(-a+b+c)(a-b+c)(a+b-c).$$

注意

$$(a+b+c)=2s,\quad (-a+b+c)=2(s-a),\quad (a-b+c)=2(s-b),\quad (a+b-c)=2(s-c),$$

因此

$$A^2=s(s-a)(s-b)(s-c),$$

从而

$$\boxed{A=\sqrt{s(s-a)(s-b)(s-c)}}.$$

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英文解析

Let angle $C$ be between sides $a$ and $b$, with the opposite side being $c$. Then the area is

$$A=\frac12 ab\sin C.$$

From the Law of Cosines

$$\cos C=\frac{a^2+b^2-c^2}{2ab}.$$

Therefore

$$\sin^2 C=1-\cos^2 C=1-\left(\frac{a^2+b^2-c^2}{2ab}\right)^2.$$

Substituting this back yields

$$A^2=\frac14 a^2 b^2\left[1-\left(\frac{a^2+b^2-c^2}{2ab}\right)^2\right] =\frac{1}{16}\left(4a^2b^2-(a^2+b^2-c^2)^2\right).$$

Expanding and simplifying the right-hand side allows it to be factored as

$$A^2=\frac{1}{16}(a+b+c)(-a+b+c)(a-b+c)(a+b-c).$$

Note that

$$(a+b+c)=2s,\quad (-a+b+c)=2(s-a),\quad (a-b+c)=2(s-b),\quad (a+b-c)=2(s-c),$$

thus

$$A^2=s(s-a)(s-b)(s-c),$$

leading to

$$\boxed{A=\sqrt{s(s-a)(s-b)(s-c)}}.$$