Prove that $X,Y,Z$ lie on a single line. 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.
Here is a simple solution using coordinates:

Consider $A = (0, 0)$, $D = (D_x, 0)$ , $B = (B_x, B_y)$ and $C = (C_x, C_y)$.

We will get $X = \left( \dfrac {C_x} 2, \dfrac {C_y} 2 \right)$ and $Y = \left( \dfrac {B_x + D_x} 2, \dfrac {B_y} 2 \right)$. After some computation we will get the coordinates $Z$ in terms of $B_x, C_x, D_x$ and $B_y, C_y, D_y$. After some simplification, the solution becomes clear (I will not provide details unless OP requests so). Then, we can just check if they are on the same line and we are done!