"Let ABCD be a square, and from base cd, let another parallelograms be created. Let's call the new parallelograms EBCF. Since it's a parallelograms, EF is equal to BC."
Something wrong there.
If If the two paralleograms have the same base, then the base CD of the square ABCD must be the same base of parallelogram EBCF. But since EBCF does not even have CD, how can the two parallelogram be on the same base?
The only "same" side on ABCD and EBCF is BC. I thought you said "...from the base cd..."
"Parallelograms which are on the same base and in the same parallels equal one another."
What that means is that if two parallelograms have the same base, and their other bases are in-line or aligned or in the same line, (or the perpendicular distance from the same base to their other bases is the same, or the two parallelograms have the same perpendicular height or altitude), then the areas of the two parallelograms are equal.
That is true.
Because whatever is the shape (or sway, or leaning) of a parallelogram, its area is always (base)*(altitude).