In $\triangle ABC$, prove that $a \sin \left(\dfrac{A}{2} + C\right) = \left(b + c\right) \sin \left(\dfrac{A}{2}\right)$
By sine rule, $\;$ $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$ $\;$ where $R$ is the circumradius of $\triangle ABC$
$\implies$ $a = 2 R \sin A$, $\;$ $b = 2 R \sin B$, $\;$ $c = 2 R \sin C$
$\begin{aligned}
RHS = \left(b + c\right) \sin \left(\dfrac{A}{2}\right) & = b \sin \left(\dfrac{A}{2}\right) + c \sin \left(\dfrac{A}{2}\right) \\\\
& = 2R \sin B \sin \left(\dfrac{A}{2}\right) + 2 R \sin C \sin \left(\dfrac{A}{2}\right) \; \left[\text{by sine rule}\right] \\\\
& = 2 R \sin \left(\dfrac{A}{2}\right) \left[\sin B + \sin C\right] \\\\
& = 2 R \sin \left(\dfrac{A}{2}\right) \times 2 \sin \left(\dfrac{B + C}{2}\right) \cos \left(\dfrac{B - C}{2}\right) \;\;\; \cdots \; (1)
\end{aligned}$
In $\triangle ABC$, $\;$ $A + B + C = \pi$
$\therefore \;$ $B + C = \pi - A$
$\therefore \;$ $\dfrac{B + C}{2} = \dfrac{\pi}{2} - \dfrac{A}{2}$
$\therefore \;$ $\sin \left(\dfrac{B + C}{2}\right) = \sin \left(\dfrac{\pi}{2} - \dfrac{A}{2}\right) = \cos \left(\dfrac{A}{2}\right)$ $\;\;\; \cdots \; (2)$
$\therefore \;$ In view of equation $(2)$, equation $(1)$ becomes
$\begin{aligned}
RHS & = 4 R \sin \left(\dfrac{A}{2}\right) \cos \left(\dfrac{A}{2}\right) \cos \left(\dfrac{B - C}{2}\right) \\\\
& = 2 R \times \left[2 \sin \left(\dfrac{A}{2}\right) \cos \left(\dfrac{A}{2}\right)\right] \times \cos \left(\dfrac{B - C}{2}\right) \\\\
& = 2 R \sin A \cos \left(\dfrac{B - C}{2}\right) \\\\
& = a \cos \left(\dfrac{B - C}{2}\right) \;\;\; \left[\text{By sine rule}\right] \;\;\; \cdots \; (3)
\end{aligned}$
$\because \;$ In $\triangle ABC$, $\;$ $A + B + C = \pi$
$\therefore \;$ $B = \pi - A - C$
$\therefore \;$ $B = \pi - A - 2 C$
$\therefore \;$ $\dfrac{B - C}{2} = \dfrac{\pi}{2} - \left(\dfrac{A}{2} + C\right)$
$\therefore \;$ $\cos \left(\dfrac{B - C}{2}\right) = \cos \left[\dfrac{\pi}{2} - \left(\dfrac{A}{2} + C\right)\right] = \sin \left(\dfrac{A}{2} + C\right)$ $\;\;\; \cdots \; (4)$
$\therefore \;$ In view of equation $(4)$ equation $(3)$ becomes
$RHS = a \sin \left(\dfrac{A}{2} + C\right) = LHS$
Hence proved.