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?
 A: Here's one way you can do it: you can use the "easier case" in which $e$ is between $b$ and $c$ and $f$ is to the right of $c$ repeatedly.  In that case you can simply cut off a triangle from $abcd$ and then reattach it with a translation to get $aefd$.  But once you've done this, you can do it again: if you have points $g$ and $h$ such that $g$ is between $e$ and $f$ and $h$ is to the right of $f$ with $gh=ef$, then repeating the same argument gives that $aefd$ and $aghd$ have the same area.  And you can repeat this over and over.
In particular, if you keep repeating this moving the top edge of the parallelogram to the right by a distance at least $bc/2$ each time, you can get it as far to the right as you want in only finitely many steps.  Thus you can eventually get that your original parallelogram has the same area as any other parallelogram with base $ad$ and whose opposite edge is on the line $bc$.
In each step of this construction, you cut the parallelogram into two pieces and then rearrange them.  So the entire process can be done in one step by cutting the original parallelogram along all the cuts you would eventually end up making, and then rearranging them in the final form.  It is complicated to see what this looks like explicitly (and the number of pieces depends on how many steps the construction takes), but it does not involve any subtraction.
A: 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
