prove That the line joining the mid-point of the diagonals of a trapezium is parallel to the parallel ..?

Question

continuation: sides and equal to half of their difference. Prove by vector method

Answer 1

I can give you an euclidean proof:

Say that the trapezium is ABCD, with AB || CD. From the midpoint M of side AD draw a parallel line to AB and CD. Suppose that it cuts the side BC at point N. According to Thales theorem BN/NC = AM/MD. But AM/MD = 1, so BN/NC = 1, so point N is the midpoint of side BC.

Now from vertex A draw a line parallel to side BC. Suppose that it cuts the median MN at point F and the side CD at point E. ABCE is a parallelogram, so EC = AB. ED = CD - EC so ED = CD - AB.
MF = ED/2, MF = (CD - AB)/2, FN = AB.
MN = MF + FN, MN = (CD - AB)/2 + AB, MN = (CD + AB)/2.

Answer 2

Not vectors again! Whatever happened to plain old geometry?