Make a sketch. 

I have my sketch before my eyes and will use it as if it is a figure. 

Let the point P be the intersection of diagonals AC and BD of the parallelogram.

The proof uses these properties from Geometry:
   
    1)  The straight line connecting midpoints of two sides of a triangle is parallel to the third side of the triangle.

    2)  If a straight line bisects one side of a triangle and is parallel to a second side of the triangle, then it bisects the third side of the triangle.

    3)  In a parallelogram, diagonals bisect each other.


    Regarding properties 1) and 2), see the lesson The line segment joining the midpoints of two sides of a triangle in this site.

    Regarding property 3), see the lesson Properties of diagonals of parallelograms in this site.


Based on Property 3), AP is half of AC.

Then AO is half of AP.

Then EO is the middle line of the triangle APD.

Hence, the line EO is parallel to DP   (Property 1).

Therefore, EF is parallel to DB.

Since EF is parallel to DB and E is the midpoint of AD, then the point F is the midpoint of AB, in accordance with the property 2).

The proof is completed.