The plane α intersects the sides AB and AC of the triangle ABC at the points M and K

The plane α intersects the sides AB and AC of the triangle ABC at the points M and K so that AM = 1 / 3AB; AK = 1 / 3AC. Prove that line BC is parallel to the plane α.

Let the length of the segment AB = X cm, AC = Y cm.

Then, according to the condition AM = AB / 3 = X / 3 cm, AK = AC / 3 = Y / 3 cm.

In triangles ABC and AMK, angle A is common.

Let’s find the ratio AB / AK and AC / AK.

X / (X / 3) = 1/3.

Y / (Y / 3) = 1/3.

Then these sides are proportional, and the triangles ABC and AMK are similar in two proportional sides and the angle between them.

Angle AMK = ABC as angles of similar triangles. But this is also the corresponding angles at the intersection of two straight lines BC and MK secant AB.

Then the BC is parallel to the CM, and since the MC belongs to the plane α, the plane itself is parallel to the BC, which was required to prove.