Solution using affine transform of the plane: send $ABCD$ to a unit square, then denote $a := AE, b := BF, c:= DG, d := AH$. Because affine transform preserves the ratio of areas the area of $EFGH$ is still $\frac{1}{2}$. Then $1$ is twice the sum of areas of 4 small right triangles near the square's angles: $ad + (1-d)c + (1-c)(1-b) + b(1-a) = 1 \Rightarrow (a-c)(d-b)=0$. Thus either $AE = DG$ or $BF = AH$. Because affine transform sends parallel lines to parallel lines and intersecting lines to intersecting lines the problem is solved. Affine transform is not really necessarily here: one can follow the same path in the original sketch, however the calculations seem to be a bit messier.