Question 1033395
Since in the Euclidean system, (absolute) distance between objects is invariant under translation/rotation/reflection of axes, we can assume that the line L in the problem is the x-axis and the parallelogram lies above the x-axis.  (If L is parallel to the x-axis, just translate the latter upward or downward; if L intersects the x-axis at point P, rotate the x-axis around the point P to correspond to line L; if the parallelogram is below the line L, reflect the former so that it is above the line L.)

Now let the leftmost vertex be A(a,b).  Then point B (assuming clockwise direction) will be (a + r, {{{b+m[1]*r}}}), where r is any positive real number and {{{m[1]}}} is the slope of the line from point A to B.

Point D will have coordinates (a+s, {{{b+m[2]*s}}}), where s again any arbitrary positive real number, while {{{m[2]}}} is the slope of the line from point A to point D.

Finally point C will have coordinates (a+r+s, {{{b+m[1]*r +m[2]*s}}}).  The line  AB, which has slope {{{m[1]}}}, will then have the same slope as line DC.

Now the distance of point A from the x-axis is b.
The distance of point B from the x-axis is {{{b+m[1]*r}}}.
The distance of point C from the x-axis is {{{b+m[1]*r +m[2]*s}}}.
The distance of point D from the x-axis is {{{b+m[2]*s}}}.

==> a+c = {{{b +b+m[1]*r +m[2]*s = 2b+m[1]*r +m[2]*s }}}.
and b + d = {{{b+m[1]*r + b+m[2]*s = 2b + m[1]*r +m[2]*s}}}

AND the result immediately follows.