Pairs of opposite sides of $\square ABCD$ meet at $E$ and $F$. Prove that the midpoint of $EF$ is collinear with the midpoints of the diagonals. 
In quadrilateral $\square ABCD$, let $AB$ and $CD$ meet at $E$, and let $AD$ and $BC$ meet at $F$. Prove that the midpoints of $AC$, $BD$, and $EF$ are collinear.

One more thing: Many times, collinearity questions can be solved by two methods, as far I know. (1) By Menelaus' theorem, or (2) by angle chasing and making the adjacent angles sum to $180^\circ$ on which three points lie. Is there any other method also? Please share.
 A: Yes, there is other method: analytic geometry.
Points $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ are collinear if and only if 
$$\begin{vmatrix}x_1 & y_1 & 1\\x_2 & y_2 & 1\\x_3 & y_3 & 1\\\end{vmatrix}=0 $$ 
Using it we can easily solve the question:
For instance,we can put the origin of oblique cartesian axes at point B such that B, A, E lie in y-axis and B, C, F lie in x-axis. Then B has coordinates $B=(0,0)$ and without loss of generality the coordinates of A, C, D, E, F can be $A=(0,2a),C=(2c,0),D=(2d_1,2d_2),E=(0,2e),F=(2f,0)$.
As points C, D, E are collinear, we get
$$\begin{vmatrix}2c & 0 & 1\\2d_1 & 2d_2 & 1\\0 & 2e & 1\\\end{vmatrix}=0, $$
$$\begin{vmatrix}c & 0 & 1\\d_1 & d_2 & 1\\0 & e & 1\\\end{vmatrix}=0. $$
On the other hand, points A, D, F are also collinear, then
$$\begin{vmatrix}0 & 2a & 1\\2d_1 & 2d_2 & 1\\2f & 0 & 1\\\end{vmatrix}=0, $$
$$\begin{vmatrix}0 & a & 1\\d_1 & d_2 & 1\\f & 0 & 1\\\end{vmatrix}=0. $$
Summing both determinants, we get
$$c(d_2-e)+a(f-d_1)+ 1(ed_1-fd_2)=0,$$
$$\begin{vmatrix}c & a & 1\\d_1 & d_2 & 1\\f & e & 1\\\end{vmatrix}=0. $$.
Therefore the points $(c,a),(d_1,d_2), (f,e)$, aka midpoints of AC, BD, EF, are collinear.
QED.
