What can say about triangle in which is
$$ b=2c\cos {\alpha} $$
Same question about triangle in which is
$$ b=2c\sin \frac {\beta}{2} $$
$$ b = 2c\sin \frac {\beta}{2} \Leftrightarrow \sin^2 \frac {\beta}{2} = \frac {b^2}{4c^2} $$
and
$$ 1- \cos \beta = 2 \sin^2 \frac {\beta}{2} = \frac {b^2}{2c^2} \ \ and \ \ \cos \beta = \frac {a^2 + c^2 - b^2}{2ac}$$
gives equality
$$ 1 - \frac {a^2 + c^2 - b^2}{2ac} = \frac {b^2}{2c^2} $$
and then to
$$ (c - a)(ac - c^2 + b^2) = 0 $$
and if
$$ ac - c^2 + b^2) \ne 0 $$
which is missing detail, then $ a = c $
Obviosly $ b < 2c \ ... $