Hey. Ive got a vectors test coming up soon, so ive been going through everything. At the moment im working on vector proofs. heres the question:(adsbygoogle = window.adsbygoogle || []).push({});

OABC is a parallelogram where [itex]\overrightarrow{OA}=\mathbf{a}[/itex] and [itex]\overrightarrow{OC}=\mathbf{c}[/itex]. M is the midpoint of the diagonal OB. Prove that the diagonals of this parallelogram bisect eachother.

I drew a diagram

http://img102.imageshack.us/img102/7223/vectorproof0rg.gif [Broken]

I already know that M is the midpoint of OB, so i just need to show that its also the midpoint of AC. This means that [itex]\overrightarrow{AM}=\overrightarrow{MC}[/itex], which is what im proving.

[tex]

\overrightarrow{OB}=\overrightarrow{OC}+\overrightarrow{CB}

[/tex]

[tex]

\overrightarrow{OB}=\mathbf{c}+\mathbf{a}

[/tex]

[tex]

\overrightarrow{AM}=\overrightarrow{AO}+\overrightarrow{OM}

[/tex]

[tex]

\overrightarrow{AM}=\mathbf{-a}+\frac{1}{2}\overrightarrow{OB}

[/tex]

[tex]

\overrightarrow{AM}=\mathbf{-a}+\frac{1}{2}(\mathbf{c+a})

[/tex]

[tex]

\overrightarrow{AM}=\frac{1}{2}(\mathbf{c-a})

[/tex]

So now i know [itex]\overrightarrow{AM}[/itex] in terms of [itex]\mathbf{a}[/itex] and [itex]\mathbf{b}[/itex].

[tex]

\overrightarrow{MC}=\overrightarrow{MO}+\overrightarrow{OC}

[/tex]

[tex]

\overrightarrow{MC}=-\frac{1}{2}\overrightarrow{OB}+\mathbf{c}

[/tex]

[tex]

\overrightarrow{MC}=-\frac{1}{2}(\mathbf{c+a})+\mathbf{c}

[/tex]

[tex]

\overrightarrow{MC}=\frac{1}{2}(\mathbf{c-a})

[/tex]

So i now also have [itex]\overrightarrow{MC}[/itex]. Since they equal the same, as required, i can conclude that:

[tex]\overrightarrow{AM}=\overrightarrow{MC}[/tex]

Is that all i would need to do to answer the question? is the information i provided sufficient to prove that the two diagonals bisect eachother?

Thanks,

Dan.

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# Vector Proofs

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