How do you prove that the midpoints of a convex quadrilateral form a parallelogram?

Answer 1

There are several different ways and these depend on the level of your knowledge of different branches of mathematics. The simplest method uses vectors, but that is probably the topic that is covered last. So this proof is based on a simpler method.

Suppose the quadrilateral is ABCD with lengths AB = 2w, BC = 2x, CD = 2y and DA = 2z (I have chosen multiples of two to avoid fractions later), and suppose that P, Q, R and S are the midpoints of the sides AB, BC, CD and DA respectively.

Consider the triangle DSR

then, by the cosine rule,

SR^2 = DS^2 + RD^2 - 2*DS*RD*cos(D) = z^2 + y^2 - 2yz*cos(D)

Now join AC and consider the triangle DAC. Again, using the cosine rule,

AC^2 = DA^2 + CD^2 - 2*DA*DC*cos(D) = 4z^2 + 4y^2 - 4*2yz*cos(D) = 4*SR^2

Taking square roots, AC = 2*SR

Therefore the sides of triangles DS are all half the lengths of the corresponding sides in triangle DAC.

Therefore the two triangles are similar and so the corresponding angles are equal.

It follows that SR is parallel to AC.

Similarly it can be shown that triangle ABC is similar to triangle PBQ so that

PQ is half of AC and PQ is parallel to AC.

Thus, SR and PQ are both parallel to AC and half its length. They are therefore parallel to one another and the same length.

Next, by drawing BD, it can be shown that QR and PS are parallel and equal.

Thus, both pairs of opposite sides of PQRS are equal and parallel which proves that the quadrilateral is a parallelogram