Prove $\triangle{ABC}$ is isosceles if $\cos A = \frac{\sin B}{2\sin C}$

Question

In $\triangle{ABC}$, $$\cos A = \frac{\sin B}{2\sin C}$$
How to prove that $\triangle{ABC}$ is isosceles?

Answer 1

Using the law of cosines $$\cos A=\frac{b^2+c^2-a^2}{2bc}$$ and the law of sines $$\sin B=\frac{b}{2R},\ \ \ \sin C=\frac{c}{2R}$$ where $R$ is the radius of the circumscribed circle of $\triangle{ABC}$, we have $$\frac{b^2+c^2-a^2}{2bc}=\frac{\frac{b}{2R}}{2\cdot\frac{c}{2R}}\iff \frac{b^2+c^2-a^2}{2bc}=\frac{b}{2c}$$$$\iff b^2+c^2-a^2=b^2\iff (c+a)(c-a)=0\iff c=a.$$

Answer 2

Using Werner Formulas, $$2\cos A\sin C=\sin(C+A)+\sin(C-A)$$

Now $\sin(C+A)=\sin(\pi-B)=\sin B$

Consequently, $\sin(C-A)=0\implies C-A=n\pi$

Now $0<A,C<\pi\implies|C-A|<\pi\implies C-A=0$

and consequently, $c=2R\sin C=2R\sin A=a$

Answer 3

You may refer to the following figure:

The three red lines are the altitudes of the triangle.

$$\cos A=\frac{\sin B}{2\sin C} \\ \implies \frac{AD}{AB}=\frac{1}{2}\frac{AE/AB}{AE/AC} =\frac{1}{2}\frac{AC}{AB} \\ \implies AD=\frac{AC}{2}$$

Then $BD$ cuts $AC$ in half. As $\triangle ADB \cong\triangle CDB$ $(SAS)$ and $BA=BC$, $\triangle ABC$ is isosceles.