In a triangle ABC, it is known that AB = BC, AM and CK is the median of this triangle. Prove that MK is parallel to AC.

Since AM and CK are the medians of the triangle, and AB = BC, then BM = BK, and then the triangles ABC and BMK are similar in two proportional sides and yy between them, and therefore the angle BAC = BMC, and since these are the corresponding angles at the intersection of straight lines AC and MK secant AB, then AC parallel to MK, which was required to be proved.