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Assume $ABCD$ is a parallelogram. $O$ is the intersection of the diagonals and $M$ an arbitrary point in the same plan. Determine $\alpha$ for which the following relation takes place:

$\overrightarrow{MA}\cdot \overrightarrow{MC} = ||OM||^2+\alpha\cdot||AC||^2$

I drew the parallelogram and chose M outside of it. How can I write the relation to get the correct result? I'm confused. Thank you very much in advance!

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up vote 1 down vote accepted

as $O$ is the intersection of the diagonals, we have $\overrightarrow{MA} = \overrightarrow{MO} + \overrightarrow{OA}$ and $\overrightarrow{MC} = \overrightarrow{MO} + \overrightarrow{OC}$. Moreover $\overrightarrow{OC}= -\overrightarrow{OA} = \frac 12\overrightarrow{AC}$. Hence \begin{align*} \overrightarrow{MA} \cdot \overrightarrow{MC} &= \bigl(\overrightarrow{MO} + \overrightarrow{OA}\bigr) \cdot (\overrightarrow{MO} + \overrightarrow{OC}\bigr)\\\ &= \|\overrightarrow{MO}\|^2 + \overrightarrow{MO} \cdot \bigl(\overrightarrow{OA} + \overrightarrow{OC}\bigr) + \overrightarrow{OA} \cdot \overrightarrow{OC}\\\ &= \|\overrightarrow{MO}\|^2 - \frac 14 \overrightarrow{AC}\cdot \overrightarrow{AC}\\\ &= \|\overrightarrow{MO}\|^2 - \frac 14 \|\overrightarrow{AC}\|^2. \\\ \end{align*} So $\alpha = -\frac 14$.

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