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.

 A: 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.
A: It is well-known that in every complete quadrilateral, the midpoints of the three diagonals are collinear. See
http://mathworld.wolfram.com/CompleteQuadrilateral.html
A: Here is an analytical solution. There are two constants $g_1,g_2$
such that $(1) \overrightarrow{QC}=g_1\overrightarrow{QB}$ and 
$(2) \overrightarrow{RA}=g_2\overrightarrow{RB}$. Denote by $L$
the intersection point of $(AD)$ with the parallel to $(DC)$ through
$B$. Then $(3) \overrightarrow{QD}=g_1\overrightarrow{QL}$ (using
Thales' theorem in triangle $QDC$) and  $(4) \overrightarrow{DA}=g_2\overrightarrow{DL}$ (using Thales' theorem in triangle $ARD$). 
Let us try to show that vectors $\overrightarrow{XZ}$ and $\overrightarrow{YZ}$ are collinear. Since $\overrightarrow{XZ}=\frac{1}{2}(\overrightarrow{CQ}+\overrightarrow{AR})=
\frac{1}{2}(g_1\overrightarrow{BQ}+g_2\overrightarrow{BR})$ and
$\overrightarrow{YZ}=\frac{1}{4}(\overrightarrow{BQ}+\overrightarrow{BR}+\overrightarrow{DQ}+\overrightarrow{DR})$, the idea is to express $\overrightarrow{DQ}+\overrightarrow{DR}$ in terms of $\overrightarrow{BQ}$ and $\overrightarrow{BR}$.
We will achieve this by finding two different relations satisfied by
$\overrightarrow{DQ}$ and $\overrightarrow{DR}$, and then solving the system. First,
note that $(5)\overrightarrow{DR}-\overrightarrow{DQ}=\overrightarrow{BR}-\overrightarrow{BQ}$.
Next, using (3) in $g_1\overrightarrow{DL}=g_1\overrightarrow{DQ}+g_1\overrightarrow{QL}$ we deduce $(6)g_1\overrightarrow{DL}=(g_1-1)\overrightarrow{DQ}$. Also,
using both (2) and (4) in $\overrightarrow{DR}=\overrightarrow{DA}+\overrightarrow{AR}$
we find that $\overrightarrow{DR}=g_2\big(\overrightarrow{DL}+\overrightarrow{BR}\big)$.
Combining this last formula with (6), we obtain
$(7) g_1\overrightarrow{DR}=g_2\big((g_1-1)\overrightarrow{DQ}+g_1\overrightarrow{BR}\big)$.
Now (5) and (7) provide us our $2\times 2$ linear system in $\overrightarrow{DQ}$ and $\overrightarrow{DR}$. Its unique solution is
$$
\overrightarrow{DQ}=\frac{g_1\overrightarrow{BQ}+(g_1g_2-g_1)\overrightarrow{BR}}{g_1+g_2-g_1g_2}, \ 
\overrightarrow{DR}=\frac{(g_1g_2-g_2)\overrightarrow{BQ}+g_2\overrightarrow{BR}}{g_1+g_2-g_1g_2}\tag{8}
$$
This yields $\overrightarrow{YZ}=\frac{1}{2(g_1+g_2-g_1g_2)}\bigg(g_1\overrightarrow{BQ}+g_2\overrightarrow{BR}\bigg)$, so that $\overrightarrow{YZ}=\frac{1}{g_1+g_2-g_1g_2}\overrightarrow{XZ}$. This concludes the proof.
A: 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!
