Proof: Heron’s Formula


Here’s the proof of Heron’s formula.

Let’s think triangle ABC. 角 \( A, B, C \) に向かい合う辺をそれぞれ \( a, b, c \) とします。

ABCbca

Now, assume that we already know triangle area is \( \frac{1}{2} bc \sin A \) .

Heron’s Formula

Triangle area \( S \) is written with \( s = a + b + c \) like below.

\[ S = \sqrt{s \left( s – a \right) \left( s – b \right) \left( s – c \right) } \]

Proof

First, present the triangle area with \( \sin \) and convert the \( sin \) representation to \( \cos \) , and 第2余弦定理を使って \( \cos \) を辺の長さのみの式に書き換えて整理すればヘロンの公式が導けます。

\begin{eqnarray*} S & = & \frac{1}{2} bc \sin A \\ & = & \frac{1}{2} bc \sqrt{ 1 – \cos ^2 A } \\ & = & \frac{1}{2} bc \sqrt{ \left( 1 – \cos A \right) \left( 1 + \cos A \right) } \end{eqnarray*}

Because of cosine formula,

\[ \cos A = \frac{b^2 + c^2 – a^2}{2bc} . \]

Therefore

\begin{eqnarray*} & & S \\ & = & \frac{1}{2} bc \sqrt{ \left( 1 – \frac{b^2 + c^2 – a^2}{2bc} \right) \left( 1 + \frac{b^2 + c^2 – a^2}{2bc} \right) } \\ & = & \frac{1}{2} \sqrt{ \left( bc – \frac{b^2 + c^2 – a^2}{2} \right) \left( bc + \frac{b^2 + c^2 – a^2}{2} \right) } \\ & = & \frac{1}{2} \sqrt{ \frac{ a^2 – b^2 + 2bc – c^2}{2} \frac{b^2 + 2bc + c^2 – a^2}{2} } \\ & = & \frac{1}{2} \sqrt{ \frac{a^2 – \left( b – c \right) ^2 }{2} \frac{ \left( b + c \right) ^2 – a^2}{2} } \\ & = & \frac{1}{4} \sqrt{ \left\{ a^2 – \left( b – c \right) ^2 \right\} \left\{ \left( b + c \right) ^2 – a^2 \right\} } \\ & = & \frac{1}{4} \sqrt{ \left(a – b + c \right) \left( a + b – c \right) \left( b + c – a \right) \left( b + c + a \right) } \\ & = & \frac{1}{4} \sqrt{ \left( 2s – 2b \right) \left( 2s – 2c \right) \left( 2s – 2a \right) 2s } \\ & = & \sqrt{ s \left( s – a \right) \left( s – b \right) \left( s – c \right) } . \end{eqnarray*}