Thread: What about triangle?

What can say about triangle in which is

$$ b=2c\cos {\alpha} $$

Same question about triangle in which is

$$ b=2c\sin \frac {\beta}{2} $$

Law of cosines

$\displaystyle a^2-c^2=b(b-2 c \cos \alpha )$

Therefore

$\displaystyle b=2 c \cos \alpha $ if and only if $\displaystyle a=c$

$$ b = 2c\sin \frac {\beta}{2} \Leftrightarrow \sin^2 \frac {\beta}{2} = \frac {b^2}{4c^2} $$

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 \ ... $

consider triangle ABC

$\displaystyle c=1$

$\displaystyle a=\sqrt{3}-1$

$\displaystyle b=\frac{-1+\sqrt{3}}{ \sqrt{2}}$

$\displaystyle b=2c \sin \left(\frac{\beta }{2}\right)$

but $\displaystyle a\neq c$

Awesome example. How you find it? BTW, problem is taken from book, so book is wrong.