In $\triangle ABC$, $\;$ $\angle A = 60^{\circ}$. Prove that $\;$ $b + c = 2 a \cos \left(\dfrac{B - C}{2}\right)$
By projection formula,
$a = b \cos C + c \cos B$, $\;$ $b = c \cos A + a \cos C$, $\;$ $c = a \cos B + b \cos A$
$LHS = b + c = c \cos A + a \cos C + a \cos B + b \cos A$ $\;\;\;$ [by projection formula]
i.e. $\;$ $b + c = \left(b + c\right) \cos A + a \left(\cos B + \cos C\right)$
i.e. $\;$ $b + c = \left(b + c\right) \cos \left(60^{\circ}\right) + 2 a \cos \left(\dfrac{B + C}{2}\right) \cos \left(\dfrac{B - C}{2}\right)$ $\;\;$ [Given: $\angle A = 60^{\circ}$]
i.e. $\;$ $b + c = \dfrac{b + c}{2} + 2 a \cos \left(60^{\circ}\right) \cos \left(\dfrac{B - C}{2}\right)$
[Note:
$\because \;$ $A$, $B$, $C$ are angles of $\triangle ABC$, $\;$ $A + B + C = 180^{\circ}$
$\therefore \;$ $B + C = 180^{\circ} - A = 180^{\circ} - 60^{\circ} = 120^{\circ}$
$\therefore \;$ $\dfrac{B + C}{2} = 60^{\circ}$]
i.e. $\;$ $\dfrac{b + c}{2} = 2 a \times \dfrac{1}{2} \times \cos \left(\dfrac{B - C}{2}\right)$
i.e. $\;$ $b + c = 2 a \cos \left(\dfrac{B - C}{2}\right)$
Hence proved.