Prove that the four lines obtained by joining each vertex of a quadrilateral to the centroid of the triangle determined by the other three vertices are concurrent.

(In reply to re: Can you provide insight? by Bractals)

For two lines we have three possibilities in Euclidean geometry:

1. the lines intersect
2. the lines are parallel *(not really a consideration given the lines we are examining for concurrency), and
3. the lines are the same line

The question arises, if the lines are the same line, do we really have two lines or are they considered one line?

If the quadrilateral were a parallelogram, AA' and CC' would be the same line, and the same would also be true for BB' and DD'.  We then have only two lines, not four.  As we need three or more lines to have concurrency, are we are missing the needed third line?

For quadrilaterals that are not parallelograms (i.e., rectangles, squares, rhombi...) there are cases where we get three and four concurrent lines, yet I am not sure we can declare concurrency when we only have two lines.

Edited on March 10, 2008, 3:09 am

Posted by Dej Mar on 2008-03-09 01:02:53