Final Round

Note that the segments $B_{1}M$ and $C_{1}M$ are medians in the right-angled triangles $B_{1}BC$ ($\angle B_1=90^{\circ }$) and $C_{1}CB$ ($\angle C_1=90^{\circ }$) respectively. So, $B_{1}M=0.5CB$ and $C_{1}M=0.5CB$, therefore $B_{1}M=C_{1}M$, i.e. $M$ is equidistant from the ends of the segment $B_1{C}_1$.

Similarly, we have $B_{1}N=0.5AH=C_{1}N$, so $N$ is equidistant from the ends of the segment $B_1{C}_1$.

Therefore $MN$ is the perpendicular bisector of the segment $B_1{C}_1$, as required.