Final answer:
By using the centroid property of a triangle and the diagonals bisecting property of a parallelogram, it can be proven that point N must be the midpoint of segment BC for the center of the parallelogram to coincide with the center of gravity of triangle MNP.
Step-by-step explanation:
To show that point N is the middle of the segment in a parallelogram ABCD, where M, N, and P are points on the sides AB, BC, and CD respectively, and the center of the parallelogram is also the center of gravity (centroid) of triangle MNP, we can use the property of the centroid of a triangle. The centroid of a triangle divides each median into two sections, with the longer section (from the vertex to the centroid) being twice as long as the shorter section (from the midpoint of the side to centroid).
In a parallelogram, the diagonals bisect each other, meaning the intersection point of the diagonals is the midpoint of each diagonal and also the center of the parallelogram. For the centroid of the triangle MNP to coincide with the center of the parallelogram, N must be at the midpoint of BC because segment NP and segment MN are medians in a triangle, and the intersection of the diagonals is the centroid.
By the definition of the centroid, which states that it divides each median into two sections in the ratio 2:1, N has to be the midpoint of BC for the center of gravity of triangle MNP to be at the center of the parallelogram.