Trapezoid Area from Slant Mid-Bridge

I took the trapezium, copied it, rotated it 180 degrees and stuck them together by the side BC:

Placing angle ABC adjacent to angle BCD gives a straight line, as the two angles are co-interior on the parallel lines AB and DC, and co-interior angles add to 180 degrees.

The resulting shape AEFD is a parallelogram, as the opposite angles are equal. This means that the left and right sides are also equal, so their midpoints would be equivalent, i.e.

$AM=MD=EO=OF$

This means that the quadrilaterals AEOM and MOFD are also parallelograms, and since opposite sides in a parallelogram are equal:

$AE=MO$

But since MO consists of two copies of MN, and AE consists of the lines AB and CD stuck together,

$AB+CD=2MN$

$=2\times 1009$

$=2018$

Now we can use the rule for the area of a trapezium:

$Area=\frac{h}{2}(a+b)$ where a and b are the parallel sides and h is the perpendicular height.

$Area=\frac{HD}{2}(AB+CD)$

$=\frac{2}{2}(2018)$

$=2018$

The solution is to turn the trapezoid into a rectangle with equal area. Draw two perpendiculars to DC, one through M and the other through N. Extend the upper horizontal AB in both directions, to intersect with the two perpendiculars and form a rectangle with length MN and height H. The two triangles “isolated” from the boundary of the rectangle, one on the left and the other on the right, are congruent with the corresponding missing triangular sections inside the boundary (similar, with equal legs H/2). Rotate the outer triangles 180 degrees about their common vertices at M and N and they will exactly fill in the missing pieces inside the boundary of the rectangle. The area of the trapezoid is equal to the area of the rectangle: 1009 x 2 = 2018. QED. I have a sketch that would have shortened the explanation, but I did not know how to insert it.