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As in the following figure of a quadrilateral;
If the diagonals are stated to bisect each other, I thought this should hold (considering the bottommost triangle in blue lines); $$\frac{\vec a+\vec b}{2}+\frac{\vec b-\vec a}{2}=\vec b$$
But this shows that all quadrilaterals have their diagonals bisecting each other, since this gives $\vec b = \vec b$, which implies it's true for all $\vec a $ and $\vec b$. Which obviously isn't true.
Where did my reasoning go wrong?,
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$\begingroup$The upper right point is not $\vec{a}+\vec{b}$ in general, if it is then we have a parallelogram, of which the diagonals bisect each other as you have proven.
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