Proving that lines from any vertex of a parallelogram to the midpoints of the opposite sides trisect a diagonal I'm asked to prove algebraically that the lines from a vertex of a parallelogram to the midpoints of the opposite sides trisect a diagonal. I did something like this

Then I got the slopes of the three lines (the two lines from the vertex and the diagonal)
The slope of the line that bisects $(a, 0)(a+b, c)$ is $\frac{c/2}{b/2 - a} = \frac{c}{b - 2a}$
The slope of the line that bisects $(0, 0)(a, 0)$ is $\frac{2c}{2b - b}$
The slope of the diagonal is ${\frac{c}{a + b}}$
Then the equations of the 3 lines are respectively
$y = \frac{c}{b -2a} (x - b) + c$ (1)
$y = \frac{2c}{2b - a} (x - b) + c$ (2)
$y = \frac{c}{a + b}x$ (3)
and then I got stuck as I would solve (1) with (3) to get the interception point and similarly (2) with (3) to get the other interception point, then prove that they trisect the diagonal. I know my way around after I get the points, but I'm not sure whether I'm on the right track. Any hints would appreciated!

The statement of the question is invariant under linear transformation. So you can assume that the parallelogram is a square and the result is trivial. Your method should work anyway, but is simpler on the square.
Syntetically you should consider one of the intersection point, let's say the upper one. Consider the two triangles given by the two lines passing through the point and the two opposite oblique sides of the parallelogram. These two triangles are similar and one is the double of the other. So you get that one part of the diagonal is half the other, and hence a third of the total.
