Question

Prove that the line segment joining the midpointsof any two sides of a triangle is parallel to the third side and equal of half of it.

Answer 1

In the triangle below, assume that D is the midpoint of AB and E is the midpoint of AC.

Then AC / AB = 1 / 2 and AD / AC = 1/2. Therefore AC / AB = AD / AC. Because angle A is congruent to itself we have ΔADE∼ΔABC by the Side-Angle-Side Similarity theorem.

This means that CD / BC = 1/2 or CD = 1/2 BC. (Which is what I think you meant by “and half of it”, i.e. the segment joining the mid-points is half of the third side of the triangle.)

Also, ∠ADE≅∠ABC because they are corresponding angles of the similar triangles. Those angles are also corresponding angles made by the transversal AB¯¯¯¯¯¯¯ that cuts DE¯¯¯¯¯¯¯ and BC¯¯¯¯¯¯¯. Because those angles are congruent, DE¯¯¯¯¯¯¯ and BC¯¯¯¯¯¯¯ must be parallel.