2. You can prove that
That means the quadrilateral EFGH must be a parallelogram.
3. The quadrilateral EFGH consists of 4 smaller parallelograms which are part of the 4 triangles which form the quadrilateral ABCD. As an example I've taken the triangle BCP and the corresponding parallelogram KFLP. Use proportions to show that the area of KFLP is of the triangle BCP.
4. You'll get this result in each of the 4 triangles. Therefore the area of the parallelogram EFGH must be of the area of the quadrilateral ABCD.