Let $ABCD$ be a parallelogram, the symmetrical point $M$ point $B$ to point $D$, and $N$ a point located on the right $BC$ like this so that $B \in (CN)$ and $BN = 2 \cdot BC$. Prove that the points $M, A, N$ are collinear.
Whether point $E$ is half of $BN$. At the bottom of the picture, you can see my ideas. Also, I thought of the propriety of the centre of gravity and that we can put a point that equals BD and then make a triangle with medians. Then we can also apply the theorem of Ceva. Hope one of you can help me! Any idea is welcome! Feel free to comment with your idea!