If $A = 45^\circ$, $B = 75^\circ$ and $C = 60^\circ$, prove that $a + c \sqrt{2} = 2b$
Given: $\;$ $A = 45^\circ$, $\;$ $B = 75^\circ$, $\;$ $C = 60^\circ$
By sine rule,
$\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = k$
where $k$ is a constant of proportionality.
$\therefore \;$ We have, $\;$ $a = k \sin A = k \sin 45^\circ = \dfrac{k}{\sqrt{2}}$ $\;\;\; \cdots \; (1a)$
$b = k \sin B = k \sin 75^\circ = \dfrac{k \left(\sqrt{3} + 1\right)}{2 \sqrt{2}}$ $\;\;\; \cdots \; (1b)$
$c = k \sin C = k \sin 60^\circ = \dfrac{\sqrt{3} k}{2}$ $\;\;\; \cdots \; (1c)$
Now, from equations $(1a)$ and $(1c)$
$\begin{aligned}
a + c \sqrt{2} & = \dfrac{k}{\sqrt{2}} + \dfrac{\sqrt{3} \times \sqrt{2} \; k}{2} \\\\
& = \dfrac{2 k + 2 \sqrt{3} \; k}{2 \sqrt{2}} \\\\
& = \dfrac{2 k \left(\sqrt{3} + 1\right)}{2 \sqrt{2}} \\\\
& = 2b \;\;\; \left[\text{in view of equation (1b)}\right]
\end{aligned}$
$\therefore \;$ $a + c \sqrt{2} = 2 b$
Hence proved.