Balkan Mathematical Olympiad

Figure 1 Clearly, $EZFY$ is a parallelogram. Let $ZY\cap AD=\{X_2\}$ and $ZY\cap BC=\{X_1\}$.

From Menelaos Theorem in triangle $ZYF$ and the line $(X_1,B,C)$ it follows $$\frac{ZB}{BF}\cdot \frac{FC}{CY}\cdot \frac{Y{X}_1}{X_{1}Z}=1,$$ hence $$\frac{X_{1}Y}{X_{1}Z}=\frac{BF}{BZ}\cdot \frac{CY}{FC}=\frac{CO}{OA}\cdot \frac{DO}{OB}.(1)$$ In similar way, from Menelaos Theorem in triangle $ZYE$ and the line $(A,D,X_2)$ we get $$\frac{X_{2}Y}{X_{2}Z}=\frac{EA}{AZ}\cdot \frac{DY}{DE}=\frac{CO}{OA}\cdot \frac{DO}{OB}.(2)$$ From (1) and (2) it follows that lines $AD$ and $BC$ intersect line $YZ$ in the same point, hence we have $X\equiv X_1\equiv X_2$. This implies the collinearity of the points $X$, $Y$, $Z$, and we are done.

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Alternative solution.

Let $M$ and $N$ be the midpoints of the sides $AD$ and $BC$, respectively, and let $Q$ be the midpoint of the segment $OP$ (Figure 2). Considering the homothecy $H_{O,\frac{1}{2}}$, it is clear that the points $E$, $F$, $P$ are collinear if and only if points $Q$, $M$, $N$ are collinear. Figure 2

Now, we know that we have the Newton-Gauss line of a complete quadrilateral for any permutation of the vertices of the quadrilaterals. In our situation we are looking for the Newton-Gauss line of the quadrilateral $ACDB$ defined by the following points: 1. $M$ the midpoint of "diagonal" $AD$, 2. $N$ the midpoint of "diagonal" $CB$, 3. $Q$ the midpoint of the segment determined by the intersections of the "opposite sides" ($AC$, $DB$) and ($AB$, $CD$)