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Hello, this problem states to prove that the line segments drawn from one vertex of the parallelogram to the midpoints of the opposite sides trisects the other diagonal. Only vector addition, subtraction, and multiplication by a scalar can be used to solve this proof. Essentially, I think you are trying to prove that A+B=3C but I can't even relate A, B, and C! Please help.

Do not post anything about proportions or side lengths or angles please, only vector addition, subtraction, and multiplication by a scalar.

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Do not post in Upper Case Letters! – Sigur Jan 15 '13 at 0:44
Just wanted to get the point across, I got rid of them. – user58304 Jan 15 '13 at 0:50
So we can't use euclidean geometry? – leo Jan 15 '13 at 1:33
nope, just basic vector geometry – user58304 Jan 15 '13 at 1:59
up vote 1 down vote accepted

You have vectors $\vec{A}$ and $\vec{B}$.

Let $C$ be the point of intersection. How is $\vec{C}$ define? It lies along the line $0 + k (\vec{A} + \vec{B})$ and also along the line $ \vec{A} + l (-2\vec{A} + \vec{B})$.

Convince yourself that $\vec{C} = \frac {1}{3} \vec{A} + \frac {1}{3} \vec{B}$. Hence, conclude that the 'mid-diagonal' trisects the diagonal.

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Why it's obvious that the point lies in the second line you claim? (it is obvious by making some drawings, but why?) – leo Jan 15 '13 at 1:37
@leo $C$ is the intersection of the main diagonal and the 'mid-diagonal'. Those lines have the respective equations, which I'm just listing out. The general vector form for a line between points $P, T$. is $\vec{P} + a (\vec{T}-\vec{P})$. In this case, it passes through the origin, and the point $A+B$. – Calvin Lin Jan 15 '13 at 2:04

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