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.



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.