The 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): $$\angle ALC=\pi -3\alpha ,\angle ALB=3\alpha ,\angle ABC=\pi -4\alpha .$$

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 $△ABL$: $\frac{b}{\sin 3\alpha }=\frac{l}{\sin 4\alpha }$. 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: $36^{\circ }$, $36^{\circ }$, $108^{\circ }$.