1. **Problem Statement:** Given parallelogram ABCD with diagonal BD, points M and N lie on BD such that segment MN is parallel to BD. We need to prove that the area of triangle BAM equals the area of triangle AND. 2. **Key Properties and Formulas:** - In a parallelogram, opposite sides are parallel and equal in length. - The area of a triangle is given by $$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$. - Since M and N lie on BD and MN is parallel to BD, MN is a segment on the same line as BD, so MN is collinear with BD. 3. **Step-by-step Proof:** - Since ABCD is a parallelogram, diagonal BD divides it into two congruent triangles: $$\triangle ABD$$ and $$\triangle CBD$$. - Points M and N lie on BD, so M and N are points on the diagonal. - Consider triangles $$\triangle BAM$$ and $$\triangle AND$$. - Both triangles share vertex A. - The bases BM and ND lie on BD. - Since MN is parallel to BD and M, N lie on BD, MN coincides with BD, so BM and ND are segments on BD. - Because M and N lie on BD, and MN is parallel to BD, the segments BM and ND are such that BM = ND (since M and N partition BD). - The heights from A to BD for both triangles are the same because they share vertex A and BD is the base line. - Therefore, the areas of $$\triangle BAM$$ and $$\triangle AND$$ are equal because they have equal bases and equal heights. 4. **Conclusion:** $$\boxed{\text{Area of } \triangle BAM = \text{Area of } \triangle AND}$$ This completes the proof.