Prove "Two parallelograms on the same base and in the same parallels, are equal." 
I understand Euclid's way of proving this. But, the book also says that I can prove this by decomposing one parallelogram into pieces, and then forming another parallelogram by combining those pieces together. 
I was thinking of dividing one parallelogram into infinitely small rectangles, and then combine them again in the contour of another parallelogram, like Riemann sum. 
But I am also assuming that this is not what author wants because calculus is not yet covered in the book. By using basic properties of parallelograms, how can you prove this postulate?

mark a series of untis lenght BC along the line BF starting at C and working to the right, and construct parallels to CD.  This divides ADEF into some set of shapes, (parallograms, triangles and and maybe an irregular quadralateral)
now do the same starting at E and working to the left.
This divides ABCD into a set of shapes.
Each shape in ABCD has a corresponding congruent shape in ADEF.
Personally, I like Euclid's proof much better.
And it is not necessary to do anything with AXD or CXE in Euclid's proof.
ABCD = ABFD - CFD and AEFD  = ABFD - BEA
