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A, B, C, D are four points in a plane with position vectors $\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}, \overline{\mathrm{d}}$ respectively such that $(\overline{\mathrm{a}}-\overline{\mathrm{d}}) \cdot(\overline{\mathrm{b}}-\overline{\mathrm{c}})=(\overline{\mathrm{b}}-\overline{\mathrm{d}}) \cdot(\overline{\mathrm{c}}-\overline{\mathrm{a}})=0$. The point $\mathrm{D}$, then is the of $\triangle \mathrm{ABC}$
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The correct answer is:
orthocentre
$\begin{aligned} & (\overline{\mathrm{a}}-\overline{\mathrm{d}}) \cdot(\overline{\mathrm{b}}-\overline{\mathrm{c}})=(\overline{\mathrm{b}}-\overline{\mathrm{d}})(\overline{\mathrm{c}}-\overline{\mathrm{a}})=0 \\ & \overline{\mathrm{AD}} \cdot \overline{\mathrm{BC}}=\overline{\mathrm{BD}} \cdot \overline{\mathrm{CA}}=0 \\ & \Rightarrow \overline{\mathrm{AD}} \perp \overline{\mathrm{BC}} \text { and } \overline{\mathrm{BD}} \perp \overline{\mathrm{CA}}\end{aligned}$
$\Rightarrow D$ is the orthocentre of $\triangle A B C$.
$\Rightarrow D$ is the orthocentre of $\triangle A B C$.
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