Belarusian Mathematical Olympiad

Let $O$ be the center of the circumscribed circle of the triangle $ABC$. Since $KBCN$ is an inscribed quadrilateral and $NB$ is the bisector of the angle $KBC$, we have $KN=NC$ (see Fig. 1). Moreover, since $N$ lies on the perpendicular bisector of the segment $AC$, we have $AN=NC$, therefore, $N$ is the circumcenter of the triangle $AKC$.

Fig. 1 Fig. 2

Construct the perpendiculars from $N$ and $O$ to the line $AB$ (see Fig. 2). Since the circumcenter is the intersection point of the perpendicular bisectors, we see that $K$ and $T$ are the feet of perpendiculars from $O$ and $N$, respectively. So, $KT=\frac{1}{2}KA=\frac{1}{2}KB$. It follows that $BK:KT=2:1$. Now by the Thales theorem, it follows that $BO:ON=BK:KT=2:1$. Thus, the circumcenter $O$ of the isosceles triangle $A_{1}B{C}_1$ coincides with the point of intersection of the medians, so this triangle is equilateral, as required.