This is the answer, in the case of a parallelogram, of a general problem, proposed by Michael Metaxas: From a point, draw two lines, dividing a given quadrangle EFGH in four parts with equal areas.
In the case of a paralellogram p = EFGH, the point must be the center (J) of it. Then there is for each direction (a), through that point, a unique other direction (b), such that the two lines intersect the parallelogram in four equal (in area) parts. The two directions are determined by the fact, that they are conjugate directions of the minimum ellipse circumscribed about the parallelogram. For a picture and a practical construction of the minimal circumscribed ellipse of the parallelogram, via the affinity mapping the square onto the parallelogram, look at the file ParaCircumscribed.html .
Look at the file TrapeziumDivision.html for a more elementary construction of the conjugate diameter (b), given the parallelogram and the direction (a).
Look at the file QuadDivision.html for the solution of the general problem of quadrangle-division, mentioned above.