59th Ukrainian National Mathematical Olympiad

Let $O$ be the intersection point of the diagonals of the parallelogram. Let's prove that the points $C_1$, $O$, $B_2$ and $K$ lie on the same circle. In fact, with the cyclicity of points $A$, $B_1$, $B_2$, $C_1$, $C_2$ and $D$ and parallelism of $AB$ and $CD$ we have (Fig. 41): $$\angle (K{C}_1,C_{1}O)=\angle (CD,DA)=\angle (BA,AD)=\angle (B_1{B}_2,B_{2}D)=\angle (B_{2}K,B_{2}O).$$ Now $\angle (BO,OK)=\angle (B_2{C}_1,C_{1}K)=\angle (B_{2}D,DC)$, thus $OK\parallel DC$. As known, diagonals $AC$ and $BD$ are divided by a point $O$ in half. By the Thales' theorem the line $OK$ (and accordingly, the point $K$) is equidistant from the lines $AB$ and $CD$.