Romanian Mathematical Olympiad

Let $a$ be the rhombus' side length. Then one can find $u,v,t\in (0,1)$ such that $AM=au$, $BN=av$, and $DP=at$.

Denote by $G$ the centroid of $MNP$ and suppose that the diagonals of the rhombus intersect at $O$. Then $G\in AC$ if and only if there exists some $k\in \mathbb{R}$ such that $\overline{OG}=k\overline{OA}$.

On the other hand, $$\begin{array}{rl}3\overline{OG}& =\overline{OM}+\overline{ON}+\overline{OP}\\ & =u\overline{OB}+(1-u)\overline{OA}+v\overline{OC}+(1-v)\overline{OB}+t\overline{OC}+(1-t)\overline{OD}\\ & =(1-u-v-t)\overline{OA}+(u+1-v-1+t)\overline{OB}\\ & =(1-u-v-t)\overline{OA}+(u-v+t)\overline{OB}.\end{array}$$ We conclude that $G\in AC$ if and only if $u-v+t=0$, which is equivalent to $AM+DP=BN$.