1. **Problem Statement:** We have a parallelogram ABCD with diagonal BD intersecting line segment AE at point F, where E is any point on side BC. We need to prove that $$\frac{DF}{EF} = \frac{FB}{FA}$$. 2. **Key Properties:** In a parallelogram, opposite sides are parallel and equal in length. Also, diagonals bisect each other. 3. **Approach:** Use similar triangles formed by the intersection and properties of parallel lines. 4. **Step 1:** Since ABCD is a parallelogram, AB is parallel to DC and AD is parallel to BC. 5. **Step 2:** Point E lies on BC, so line AE intersects BD at F. 6. **Step 3:** Consider triangles $\triangle DFE$ and $\triangle BFA$. 7. **Step 4:** Because AB is parallel to DC, angles $\angle DFE$ and $\angle BFA$ are equal (alternate interior angles). 8. **Step 5:** Also, angles $\angle DEF$ and $\angle FAB$ are equal (corresponding angles due to parallel lines). 9. **Step 6:** By AA similarity criterion, $\triangle DFE \sim \triangle BFA$. 10. **Step 7:** From similarity, corresponding sides are proportional: $$\frac{DF}{FB} = \frac{EF}{FA}$$ 11. **Step 8:** Rearranging gives: $$\frac{DF}{EF} = \frac{FB}{FA}$$ **Final answer:** $$\boxed{\frac{DF}{EF} = \frac{FB}{FA}}$$