If E is the midpoint of the segment BC, L 1 ∥ L 2 ,...

Step 1
Draw a line parallel to AD and passing through E. Suppose that that line intersects L1 and L2 at $C^{\prime }$ and $B^{\prime }$, respectively.
The triangle AED has half the area of the parallellogram $A{B}^{\prime }{C}^{\prime }D$ .
The parallelogram $A{B}^{\prime }{C}^{\prime }D$ has the same area as the trapezoid because the triangle $BE{B}^{\prime }$ and $CE{C}^{\prime }$ are equal (ALA).
Therefore the triangle AED has half the area of the trapezoid ABCD too.

Step 1
Call the perpendicular distance between the lines k. Then the area of $\mathrm{△}CDE$ is $\frac{1}{2}CD\frac{k}{2}=\frac{CDk}{4}.$ . The area of $\mathrm{△}AEB$ is $\frac{1}{2}AB\frac{k}{2}=\frac{ABk}{4}.$ . The area of the trapezoid ABCD is $k\frac{CD+AB}{2}$ .
Comparing these areas shows that $\mathrm{△}ADE$ is half the area of the trapezoid.