Problems from Ukrainian Authors

Let $D^{\prime }$ be symmetric to $D$ with respect to $AC$. If $D^{\prime }$ coincides with $B$, then $N$ and $M$ are symmetric to $K$ and $L$, respectively, with respect to $AC$, and the problem clearly holds. Let $D^{\prime }\ne B$. Since ${\omega }_2$ is symmetric to ${\omega }_1$ with respect to $AC$, If we denote by $M^{\prime }$ and $N^{\prime }$ the points symmetric to $M$ and $N$, respectively, with respect to $AC$ we will have that $B,D^{\prime },M^{\prime },L,N^{\prime },K$ all lie on ${\omega }_1$ (fig. 24). Since $\angle A{D}^{\prime }C=\angle ADC=\angle ABC$, points $A,B,D^{\prime },C$ are concyclic. We get $\angle (CA,A{D}^{\prime })=\angle (CB,B{D}^{\prime })=\angle (L{N}^{\prime },N^{\prime }{D}^{\prime })$, so $L{N}^{\prime }\parallel AC$, and similarly $M^{\prime }K\parallel AC$. Then $L{M}^{\prime }K{N}^{\prime }$ is a cyclic trapezoid, so it's isosceles, and $KL=M^{\prime }{N}^{\prime }=MN$. We also get $\angle (MN,AC)=\angle (AC,M^{\prime }{N}^{\prime })=\angle (N^{\prime }L,M^{\prime }{N}^{\prime })=\angle (KL,N^{\prime }L)=\angle (KL,AC)$, so $MN\parallel KL$. Since segments $KL$ and $MN$ are equal, $KLMN$ is a parallelogram, so $KM$ intersects $NL$ at its midpoint, and since $N^{\prime }L\parallel AC$ and $N$ is symmetric to $N^{\prime }$ with respect to $AC$, the midpoint of $NL$ lies on $AC$, as desired.