In triangle ABC, it is known that AB = BC, AM and CK are the medians of this triangle. Prove that MK || AC.

Since points M and K are the middle of the segments BC and AB, then

ВK / AВ = 1/2, ВM / ВС = 1/2.

Then the triangles ABC and ВKM are similar in two proportional sides and the angle between them, since they have a common angle B.

Then the angle ВKM = BAC, and since these are the corresponding angles for straight lines KM and AC and secant AB, the segment KM is parallel to the base of AC, which was required to be proved.