Let $\gamma$ be the circumcircle of the acute triangle $ABC$. Let $P$ be the midpoint of the minor arc $BC$. The parallel to $AB$ through $P$ cuts $BC, AC$ and $\gamma$ at points $R,S$ and $T$, respectively. Let $K \equiv AP \cap BT$ and $L \equiv BS \cap AR$. Show that $KL$ passes through the midpoint of $AB$ if and only if $CS = PR$.
Since $ABPT$ is an isosceles trapezium we have that $K$ is in the perpendicular bisector of $AB$.
It is easy to see, angle chasing and using the fact that $BP = CP$, that $CS = PR$ implies triangles $BRP$ and $PSC$ being congruent. From this, and easy angle chase it follows that $ABC$ is isosceles with $CB=CA$. Hence $ASRP$ is also an isosceles trapezium and therefore $L$ is on the perpendicular bisector of $AB$. The other direction is the same argument...
It can be proved by similarity and power of point that if $CS=PR$ then triangle $ABC$ is isosceles and vice-versa. Then it's easy to show that $C, L, K, M$ are collinear and it concludes it.