Problems of Ukrainian Authors

Let's find an isosceles triangle. Let's pretend that the vertex angle is obtuse. Then let's denote the base angle as $2\alpha$ (Fig. 49). Then we can easily find values of some angles (Fig. 49):
$$\angle ALC = \pi - 3\alpha, \quad \angle ALB = 3\alpha, \quad \angle ABC = \pi - 4\alpha.$$

Let's denote the bisection and the sides as follows: $AL = l$, $AD = h$, $AB = b$, then $2h = l$.
Then $h = b \sin 2\alpha$. By the law of sines for $\triangle ABL$: $\frac{b}{\sin 3\alpha} = \frac{l}{\sin 4\alpha}$. So, $2h = 2b \sin 2\alpha = \frac{b \sin 4\alpha}{\sin 3\alpha}$
$\Rightarrow 2 \sin 2\alpha \sin 3\alpha = 2 \sin 2\alpha \cos 2\alpha \Rightarrow \sin 3\alpha = \cos 2\alpha$.
The answer of this equality is pretty clear: $\alpha = 18^\circ$.

Answer: $36^\circ$, $36^\circ$, $108^\circ$.