{%hackmd @RintarouTW/About %}

# Heron's Formula

<iframe src="https://rintaroutw.github.io/fsg/test/HeronFormula.svg" width="883" height="625"></iframe>

$$
\cases{
b^2=h^2+d^2\\
a^2=h^2+(c-d)^2
}\\
a^2-b^2=c^2-2cd\\
d=\frac{b^2+c^2-a^2}{2c}
$$

:::info
We can also get the same result directly from $a^2=b^2+c^2-2bc\cos\theta$, where $d=b\cos\theta$
$$
\implies a^2=b^2+c^2-2cd\\
d=\frac{b^2+c^2-a^2}{2c}
$$
:::

$$
\begin{array}l
h^2&=b^2-d^2\\
&=(b+d)(b-d)\\
&=(b+\frac{b^2+c^2-a^2}{2c})(b-\frac{b^2+c^2-a^2}{2c})\\
&=(\frac{(b+c)^2-a^2}{2c})(\frac{a^2-(b-c)^2}{2c})\\
&=(\frac{(b+c+a)(b+c-a)}{2c})(\frac{(a+b-c)(a-b+c)}{2c})
\end{array}\\
\therefore h = \frac{\sqrt{(b+c+a)(b+c-a)(a+b-c)(a-b+c)}}{2c}\\
$$

$$
\begin{array}l
\triangle &= \frac{ch}{2}\\
&= \frac{c}{2}\frac{\sqrt{(b+c+a)(b+c-a)(a+b-c)(a-b+c)}}{2c}\\
&=\frac{\sqrt{(b+c+a)(b+c-a)(a+b-c)(a-b+c)}}{4}\\
&=\sqrt{\frac{(b+c+a)(b+c-a)(a+b-c)(a-b+c)}{16}}\\
&=\sqrt{\frac{b+c+a}{2}\frac{b+c-a}{2}\frac{a+b-c}{2}\frac{a-b+c}{2}}\\
&=\sqrt{s(s-a)(s-b)(s-c)}
\end{array}
$$

## References

- https://en.wikipedia.org/wiki/Heron%27s_formula
- http://jwilson.coe.uga.edu/EMT668/EMAT6680.2000/Umberger/MATH7200/HeronFormulaProject/GeometricProof/geoproof.html

###### tags: `math` `geometry` `Heron`