Why?
The leftmost drawing below is a trapezoid. Draw vertical segments from the top base to the bottom through the midpoints of the left and right sides and draw another segment connecting those two midpoints; then connect the end of each vertical segment that lies outside the trapezoid to the closest vertex of the trapezoid (see the red segments in the rightmost drawing).
Notice that on each side of the rightmost figure there are two CONGRUENT (why congruent?) triangles, one outside the trapezoid and one inside. If you take the two inside-triangles off and put them where the respective outside triangles are, then the modified figure has the same area as the original trapezoid. But, by the way we constructed the modified figure, it is a rectangle! The height of the rectangle is h, the height of the trapezoid. The width of the rectangle is the length of segment joining the side-midpoints which is the average of the two bases and (why is it the average?) So the area of the trapezoid is the area of the rectangle, which is the height of the trapezoid times the average of the two bases:
(
