Introduction:
In geometry, a parallelogram is a quadrilateral with opposite sides that are parallel and equal in length. In this proof, we will show that parallelograms on the same base and between the same parallels are equal in area.

Proof:
Let's consider two parallelograms ABCD and PQRS, both on the same base AB and between the same parallels AB and CD.

Step 1: Construct Diagonal AC:
Draw the diagonal AC of parallelogram ABCD. Since opposite sides of a parallelogram are parallel and equal in length, we know that AD is parallel to BC and AB is parallel to CD.

Step 2: Triangles ACD and BCD:
We can divide parallelogram ABCD into two triangles, ACD and BCD, by drawing diagonal AC.

Step 3: Proving ΔACD ≅ ΔBCD:
Since AD is parallel to BC, and AC is a transversal, we have alternate interior angles ACD and BCD that are congruent. Additionally, both triangles share the common side CD. Therefore, by the Side-Angle-Side (SAS) congruence criterion, we can conclude that ΔACD ≅ ΔBCD.

Step 4: Area of Triangles:
The area of a triangle is given by the formula A = 1/2 * base * height. Since both triangles have the same base CD, their heights will be equal. Thus, the areas of triangles ACD and BCD are equal.

Step 5: Area of Parallelograms:
The area of a parallelogram is given by the formula A = base * height. Since the base of parallelograms ABCD and PQRS is AB, and the height is the same for both parallelograms (as shown in Step 4), we can conclude that the areas of parallelograms ABCD and PQRS are equal.

Conclusion:
Therefore, we have proved that parallelograms on the same base and between the same parallels are equal in area.