Let O be the origin and let PQR be an arbitra | vedclass

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(D) Given the equation: $\overline{OP} \cdot \overline{OQ} + \overline{OR} \cdot \overline{OS} = \overline{OR} \cdot \overline{OP} + \overline{OQ} \cd

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orthocenter

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(D) Given the equation: $\overline{OP} \cdot \overline{OQ} + \overline{OR} \cdot \overline{OS} = \overline{OR} \cdot \overline{OP} + \overline{OQ} \cdot \overline{OS}$.Rearranging the terms: $\overline{OP} \cdot \overline{OQ} - \overline{OR} \cdot \overline{OP} = \overline{OQ} \cdot \overline{OS} - \overline{OR} \cdot \overline{OS}$.$\overline{OP} \cdot (\overline{OQ} - \overline{OR}) = \overline{OS} \cdot (\overline{OQ} - \overline{OR})$.$(\overline{OP} - \overline{OS}) \cdot (\overline{OQ} - \overline{OR}) = 0$.$\overline{SP} \cdot \overline{RQ} = 0$.This implies $\overline{SP} \perp \overline{RQ}$,meaning $S$ lies on the altitude from $P$ to $QR$.Similarly,from $\overline{OR} \cdot \overline{OP} + \overline{OQ} \cdot \overline{OS} = \overline{OQ} \cdot \overline{OR} + \overline{OP} \cdot \overline{OS}$,we get $\overline{SQ} \perp \overline{PR}$.Since $S$ is the intersection of altitudes,$S$ is the orthocenter of $	riangle PQR$.

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