Cosine Rule (Proof)

Proof:

Case 1: \(\angle A = 90^\circ \)

Since
$$\begin{align}
\cos A & = \cos 90^\circ \\
& = 0 \\
\end{align}$$
Then,
\[2bc\cos A = 0\]
By Pythagoras' Theorem,
$$\begin{align}
{a^2} & = {b^2} + {c^2} \\
& = {b^2} + {c^2} - 2bc\cos A \\
\end{align}$$

Case 2: \(\angle A < 90^\circ \)

Let \(CP = h\) be the perpendicular height of \(\Delta ABC\).
Let \(AP = m\) and \(BP = n\) such that \(n = c - m\).
$$\begin{align}
\cos A & = \frac{m}{b} \\
m & = b\cos A \\
\end{align}$$
From \(\Delta ACP\), by Pythagoras' Theorem,
\[{h^2} = {b^2} - {m^2}\]
From \(\Delta BCP\), by Pythagoras' Theorem,
\[{h^2} = {a^2} - {\left( {c - m} \right)^2}\]
Therefore,
$$\begin{align}
{b^2} - {m^2} & = {a^2} - {\left( {c - m} \right)^2} \\
& = {a^2} - {c^2} + 2cm - {m^2} \\
{b^2} & = {a^2} - {c^2} + 2cm \\
{a^2} & = {b^2} + {c^2} - 2cm \\
& = {b^2} + {c^2} - 2bc\cos A \\
\end{align}$$

Case 3: \(\angle A > 90^\circ \)

Let \(CP = h\) be the perpendicular height of \(\Delta ABC\).
Let \(AP = m\).
$$\begin{align}
\cos A & = - \cos \left( {180^\circ - A} \right) \\
& = - \frac{m}{b} \\
m & = - b\cos A \\
\end{align}$$
From \(\Delta ACP\), by Pythagoras' Theorem,
\[{h^2} = {b^2} - {m^2}\]
From \(\Delta BCP\), by Pythagoras' Theorem,
\[{h^2} = {a^2} - {\left( {c + m} \right)^2}\]
Therefore,
$$\begin{align}
{b^2} - {m^2} & = {a^2} - {\left( {c + m} \right)^2} \\
& = {a^2} - {c^2} - 2cm - {m^2} \\
{b^2} & = {a^2} - {c^2} - 2cm \\
{a^2} & = {b^2} + {c^2} + 2cm \\
& = {b^2} + {c^2} - 2bc\cos A \\
\end{align}$$