In $\triangle ABC$, prove that $2 a c \sin \dfrac{1}{2} \left(A - B + C\right) = c^2 + a^2 - b^2$
In $\triangle ABC$, $\;$ $A + B + C = \pi$ $\implies$ $A + C = \pi - B$
$\begin{aligned}
LHS & = 2 a c \sin \left(\dfrac{A - B + C}{2}\right) \\\\
& = 2 a c \sin \left(\dfrac{\pi - B - B}{2}\right) \\\\
& = 2 a c \sin \left(\dfrac{\pi}{2} - B\right) \\\\
& = 2 a c \cos B \\\\
& = 2 a c \left(\dfrac{c^2 + a^2 - b^2}{2 c a}\right) \;\; \left[\text{by cosine rule}\right] \\\\
& = c^2 + a^2 - b^2 = RHS
\end{aligned}$
Hence proved.