Prove that $X,Y,Z$ lie on a single line. [duplicate]

Let $ABCD$ be a convex quadrilateral such that no two opposite sides are parellel to each other. Denote by $Q$ the intersection of lines $AD$ and $BC$ and by $R$ the intersection of lines $AB$ and $CD$. Let $X,Y,Z$ be midpoints of $AC, BD$ and $QR$ respectively. Prove that $X,Y,Z$ lie on the same line.

I am not getting any approach to solve this question. Please help.

This is also known the Newton-Gauss line of $ABCD$. The usual proof is either by area considerations or by Menelaus' theorem. However, I present a purely synthetic approach.

Let $L,M,N$ be the midpoints of $\overline{AB},\overline{AR},\overline{BR}$. By considering midpoints in $\triangle ABC$, we have $XL\parallel BC$, and with $\triangle BQR$, we have $BQ\parallel NZ\implies XL\parallel NZ$. In a similar way, we can obtain $LY\parallel MZ$ and $MX\parallel NY$. But now Pappus' theorem on the hexagon $XMZNYL$ implies that $X,Y,Z$ are collinear.

It is well-known that in every complete quadrilateral, the midpoints of the three diagonals are collinear. See http://mathworld.wolfram.com/CompleteQuadrilateral.html