Let $OP$ and $OQ$ be the perpendiculars from the circumcenter $O$ of a triangle $ABC$ to the internal and external bisectors of the angle $B$. Prove that the line$ PQ$ divides the segment connecting midpoints of $CB$ and $AB$ into two equal parts.
(Artemiy Sokolov)
Let the midpoints of $BC$ be $M$ and $AB$ be $N$. Then since $\angle OQB=\angle OPB=\angle OMB=\angle ONB=\dfrac{\pi}{2}$ Therefore, $OPNBQM$ is a cyclic hexagon with diameter $OB$.However $OPBQ$ is a rectangle hence $PQ$ is also a diameter. It suffices to show that $PQ \perp MN$. But $P,Q$ are the midpoints of the segments joining the arc midpoints to vertex $B$ and hence $PQ$ is $\perp$ to $AC$.However $MN || AC$ so we are done.