Let γ 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 γ at points R,S and T, respectively. Let K≡AP∩BT and L≡BS∩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.