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,
$$\begin{align} {b^2} - {m^2} & = {a^2} - {\left( {c + m} \right)^2} \\ & = {a^2} - {c^2} - 2cm - {m^2} \\ {b^2} & = {a^2} - {c^2} - 2cm \\ {a^2} & = {b^2} + {c^2} + 2cm \\ & = {b^2} + {c^2} - 2bc\cos A \\ \end{align}$$