# Thread: vecror problem for the best only

1. ## vecror problem for the best only

ps : sorry for my english ; i don't speak english very well it obvious .so sorry

2. ## Re: vecror problem for the best only

Tedious, but I think the most straightforward way is to set up a coordinate system so that B is at the origin and C is on the x-axis, at (c, 0). You can take A to be at (a, b) and then find the coordinates of the other points.

3. ## Re: vecror problem for the best only

Originally Posted by hacen
As written (or labelled) I don't think it is true.

Consider that $\overleftrightarrow {DE} \cap \overleftrightarrow {CE} = \left\{ E \right\}$, but if $G \in \overleftrightarrow {DE} \wedge G \in \overleftrightarrow {CE}$ then that means
$G=E$ but $G$ is the mid-point of $\overline{CE}$.

4. ## Re: vecror problem for the best only

sorry it my fault ;it is not line (DE) but is (DB)
so : Prove that lines (DB) and (AF) met in G.

5. ## Re: vecror problem for the best only

Put A at (0,0), B at (0,3), C at (4,0).
According to the problem statement, D is at (3,0), E is at (0,2) and G (midpoint of CE) is at (2,1).
x - 2y = 0 defines the line AG.
3x + 4y = 12 defines the line BC.
Solve for the intersection and you get F at (12/5,6/5) which is 2/5 of the way from C to B.
So AF and CE intersect at G for the 3-4-5 triangle with A at the origin.
Stretching the axes and varying the angle between them preserves the proportional placement of the points and demonstrates the relation for any triangle.