The idea is to prove the cosines law: $$ a^2 = b^2 + c^2 - 2bc\cos\theta $$ where the variables are defined by the triangle: $$ \begin{xy} ,(0,0) ;(40,0)**@{-} ;(60,30)**@{-} ;(0,0)**@{-} ,(20,-3)*{c} ,(7,2)*{\theta} ,(50,12)*{a} ,(30,17)*{b} \end{xy} $$

Let's add a couple of lines and two variables, to get $$ \begin{xy} ,(0,0) ;(40,0)**@{-} ;(60,30)**@{-} ;(0,0)**@{-} ,(20,-3)*{c} ,(7,2)*{\theta} ,(50,12)*{a} ,(30,17)*{b} ,(40,0) ;(60,0)**@{--} ;(60,30)**@{--} ,(50,-3)*{x} ,(63,15)*{y} \end{xy} $$

This is all we need. We can use Pythagoras' theorem to show that $$ a^2 = x^2 + y^2 $$ and $$ b^2 = y^2 + \left(c+x\right)^2 $$

So, combining these two we get \begin{eqnarray*} a^2 & = & x^2 + b^2 - \left(c+x\right)^2\\ a^2 & = & x^2 + b^2 - c^2 - 2cx - x^2\\ a^2 & = & b^2 - c^2 - 2cx \end{eqnarray*} So, all we need now is an expression for $x$ Well, we can use the definition of the cosine function to show that \begin{eqnarray*} c + x & = & b \cos\theta\\ x & = & b\cos\theta - c \end{eqnarray*} With this result in hand, we find that \begin{eqnarray} a^2 & = & b^2 - c^2 - 2cx\nonumber\\ a^2 & = & b^2 - c^2 - 2c\left( b\cos\theta - c\right)\nonumber\\ a^2 & = & b^2 - c^2 -2bc\cos\theta + 2c^2\nonumber\\ a^2 & = & b^2 + c^2 - 2bc\cos\theta \end{eqnarray}