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$a,b,c$为三角形三边长，$p = \frac{1}{2}\left( {a + b + c} \right)$，则三角形面积 $S = \sqrt {p\left( {p - a} \right)\left( {p - b} \right)\left( {p - c} \right)} $
$S = \frac{1}{2}ab\sin C = \frac{1}{2}ab\sqrt {1 - \cos ^2 C} $
在△ABC中，由余弦定理可得：
$\cos C = \frac{{a^2 + b^2 - c^2 }}{{2ab}}$
代入上式可得
$S = \frac{1}{2}ab\sqrt {1 - \frac{{\left( {a^2 + b^2 - c^2 } \right)^2 }}{{4a^2 b^2 }}} = \frac{1}{2}ab\sqrt {\frac{{4a^2 b^2 - \left( {a^2 + b^2 - c^2 } \right)^2 }}{{4a^2 b^2 }}} $
$ = \sqrt {\frac{{4a^2 b^2 - \left( {a^2 + b^2 - c^2 } \right)^2 }}{{16}}} = \sqrt {\frac{{\left( {2ab + a^2 + b^2 - c^2 } \right)\left( {2ab - a^2 - b^2 + c^2 } \right)}}{{16}}} $
$ = \sqrt {\frac{{\left[ {\left( {a + b} \right)^2 - c^2 } \right]\left[ {c^2 - \left( {a - b} \right)^2 } \right]}}{{16}}} = \sqrt {\frac{{\left( {a + b + c} \right)\left( {a + b - c} \right)\left( {c + a - b} \right)\left( {c - a + b} \right)}}{{16}}} $
又$p = \frac{1}{2}\left( {a + b + c} \right)$ ， 故$\frac{{a + b - c}}{2} = \frac{{2p - 2c}}{2} = p - c$ ，$\frac{{c + a - b}}{2} = \frac{{2p - 2b}}{2} = p - b$ ，$\frac{{c - a + b}}{2} = \frac{{2p - 2a}}{2} = p - a$ ，故$S = \sqrt {p\left( {p - a} \right)\left( {p - b} \right)\left( {p - c} \right)} $．