In the triangle ABC, M is the midpoint of AB and N is the midpoint of AC. Prove that CM and BN cannot bisect.
(I can prove it by contradiction if it is a scalene triangle, but i cant seem to prove it for isosceles)
I am not sure what 'bisect' means here. Here are two possible ways to look at the problem:
If CM and BN meet in a common midpoint, then CNMB would be a parallelogram (that is, CN would have to be parallel to BM).
Two medians CM and BN of a triangle meet in a point G (the centroid of the triangle) that lies 2/3 of the way from the vertex to the midpoint opposite side (so that CG = 2GM and BG = 2GN).
Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.