NBlog

由一条咸鱼搭建的博客

证明海伦公式

By NriotHrreion2023-02-04

很早就想尝试了, 今天就来挑战一下吧!

什么是海伦公式?

设平面内有一个三角形ABC, 其三边边长分别为a、b、c, 则该三角形的面积可以由下面的公式求得:

S = p (p - a) (p - b) (p - c)

其中, p为三角形周长的一半, 即

p = \frac{a + b + c}{2}

证明

设平面内有一个三角形ABC, 其三边边长分别为AB=a、BC=b、AC=c

作AD⊥BC, 设AD=h, CD=x

由勾股定理, 得

c^{2} - x^{2} = h^{2}

B D = a^{2} - h^{2}

因为

b - B D = C D = x

所以

x = b - a^{2} - h^{2}

b - x = a^{2} - h^{2}

b^{2} - 2 b x + x^{2} = a^{2} - h^{2}

得

h^{2} = a^{2} - b^{2} + 2 b x - x^{2}

又因为

h^{2} = c^{2} - x^{2}

所以

a^{2} - b^{2} + 2 b x - x^{2} = c^{2} - x^{2}

a^{2} - b^{2} + 2 b x = c^{2}

2 b x = c^{2} - a^{2} + b^{2}

得

x = \frac{c ^{2} - a ^{2} + b ^{2}}{2 b}

将其代入 $h^{2} = c^{2} - x^{2}$ 中, 得

h = c^{2} - (\frac{c ^{2} - a ^{2} + b ^{2}}{2 b})^{2}

接着, 由三角形面积公式, 得

S = \frac{1}{2} b c^{2} - (\frac{c ^{2} - a ^{2} + b ^{2}}{2 b})^{2}

化简:

S = \frac{b ^{2}}{4} (c + \frac{c ^{2} - a ^{2} + b ^{2}}{2 b}) (c - \frac{c ^{2} - a ^{2} + b ^{2}}{2 b})

= \frac{b ^{2}}{4} (\frac{2 c b + c ^{2} - a ^{2} + b ^{2}}{2 b}) (\frac{2 c b - c ^{2} + a ^{2} - b ^{2}}{2 b})

= \frac{b ^{2}}{4} [\frac{( b + c ) ^{2} - a ^{2}}{2 b}] [\frac{a ^{2} - ( b - c ) ^{2}}{2 b}]

= \frac{1}{4} \times \frac{( b + c + a ) ( b + c - a )}{2} \times \frac{( a + b - c ) ( a - b + c )}{2}

= \frac{a + b + c}{2} \times \frac{b + c - a}{2} \times \frac{a + b - c}{2} \times \frac{a - b + c}{2}

= \frac{a + b + c}{2} \times \frac{a + b + c - 2 a}{2} \times \frac{a + b + c - 2 c}{2} \times \frac{a + b + c - 2 b}{2}

= \frac{a + b + c}{2} \times (\frac{a + b + c}{2} - a) (\frac{a + b + c}{2} - c) (\frac{a + b + c}{2} - b)

即

S = p (p - a) (p - b) (p - c)

其中

p = \frac{a + b + c}{2}

证毕.