# 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.

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.