Given a triangle ABC, the alpha plane is drawn through the midpoints of its sides AB and BC.

Given a triangle ABC, the alpha plane is drawn through the midpoints of its sides AB and BC. Prove that AC is parallel to the alpha plane.

Let the plane intersect AB and BC at points M and N, respectively. Triangle MBN is similar to triangle ABC: angle B is common, sides MB and NB are proportional to sides AB and BC by a factor of 1/2.
For similar triangles, the corresponding angles are: <BAC = <BMN. The indicated angles are one-sided angles corresponding when cutting the segments MN and AC by the segment AB. It follows from this that AC is parallel to MN. MN belongs to the plane a (if the points M and N of the segment belong to the plane, then all the points of the segment also lie in this plane). AC is parallel to plane a because it is parallel to the segment MN belonging to plane a.