Urgent geometry help is needed here. There's a problem on triangles that I just cannot get!?

Question

OK, here's the question:
A line from vertex C of triangle ABC bisects the median from A, meeting AB at E. Prove that E divides AB in the ratio 1 : 2.

Answer 1

here is a vector approach to the solution
LET vector BC = t
vector BA = u
CE = CB + BE = -t + k u, where k is a constant to be determined
AM = AB + BM = -u + 0.5 t
CO = -0.5 t + 0.5( u - 0.5t) = 0.5 u - 0.75 t (where O is the midpoint of AM and M the midpoint of BC.
Now, since c, O, E are collinear, the vectors CO and CE are parallel, hence
k /-1 = 0.5/ -0.75
k = 2/3
this says that BE = 2/3 BA, which means EA = 1/3 BA
BE: EA = (2/3): (1/3) = 2: 1
or EA : BE = 1 :2
QED.
I don't know if you took vectors, but it makes the solution of those types of questions simple.
Here is another solution in case you do not know vectors.
Let M be midpoint of BC, O the point of intersection of CE and AB and through M draw the median (MT) to AC, intersecting CE at P.
IN triangle AMC, MT and CO are medians hence MP/ PT = 2/1 = BE/ EA (why?) you convince yourself of this.
thus EA: BE = 1: 2

Answer 2

Definition of a median
Median of a triangle is a line that is parallel to the base and is 1/2 length with the base

If you have now the triangle ABC and DE is the median of the triangle . . . it is given
DE = 1/2 BC . . . is given
DE // BC . . . . .is given
Triangle ADE is congruent to triangle ABC
DE / BC = AE / AB . . . . side of congruent triangles are proportional
DE / BC = 1/2 . . . definition of median
therefore AE / AB = 1/2 . . . . by substitution

Answer 3

Let A' be the middle of BC. Draw the parallel to your line, going through A' and conclude with Thales.

Answer 4

Not enough info.