1. **Problem Statement:** Given two parallelograms ABCD and ABEF standing on the same base AB and between the same parallel lines AB and FC, prove that the area of ABCD equals the area of ABEF. 2. **Key Concept:** The area of a parallelogram is given by the formula: $$\text{Area} = \text{base} \times \text{height}$$ where the height is the perpendicular distance between the two parallel lines containing the base and the opposite side. 3. **Explanation:** Since ABCD and ABEF share the same base AB, the length of AB is the same for both parallelograms. 4. Both parallelograms lie between the same parallel lines AB and FC, so the perpendicular distance (height) between these lines is the same for both. 5. Therefore, the area of ABCD is: $$\text{Area}_{ABCD} = AB \times h$$ 6. Similarly, the area of ABEF is: $$\text{Area}_{ABEF} = AB \times h$$ 7. Since both areas are calculated using the same base and height, it follows that: $$\text{Area}_{ABCD} = \text{Area}_{ABEF}$$ **Final answer:** $$\boxed{\text{Area of } ABCD = \text{Area of } ABEF}$$