Attachment 28743

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

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- Jul 6th 2013, 03:56 AMhacenvecror problem for the best only
Attachment 28743

ps : sorry for my english ; i don't speak english very well it obvious (Speechless) .so sorry (Cool) - Jul 6th 2013, 04:28 AMHallsofIvyRe: 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.

- Jul 6th 2013, 04:51 AMPlatoRe: vecror problem for the best only
As written (or labelled) I don't think it is true.

Consider that $\displaystyle \overleftrightarrow {DE} \cap \overleftrightarrow {CE} = \left\{ E \right\}$, but**if**$\displaystyle G \in \overleftrightarrow {DE} \wedge G \in \overleftrightarrow {CE}$ then that means

$\displaystyle G=E$ but $\displaystyle G$ is the mid-point of $\displaystyle \overline{CE}$.

**Have I miss-read the problem?** - Jul 6th 2013, 08:02 AMhacenRe: vecror problem for the best only
sorry it my fault ;it is not line (DE) but is (DB)

so : Prove that lines) and (AF) met in G.__(DB__

Attachment 28745 - Aug 14th 2013, 05:52 PMPaddyMacRe: 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.