prove That the line seg 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 think you mean a trapezoid

Answer 2

Every trapezoid can be converted by scaling, translation, rotation into a trapezoid with vertices

A:(a, -1); B:(b, 1); C:(c, 1); D:(d, -1)

where a < d and b < c.

(the parallel sides AD and BC are horizontal)

The midpoints of the diagonals are the averages of opposite vertices:

of AC: P:([a+c]/2, 0)
of BD: Q:([b+d]/2, 0)

The connecting line PQ is the x-axis, and therefore parallel to the horizontal sides.

The length of segment PQ is the absolute difference

L = |[b+d]/2 - [a+c]/2| = |d - a + b - c|/2

The difference between the lengths AD and CD is

D = |(d - a) - (c - b)| = |d - a + b - c|

and it is clear that L = D/2