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ABC is a triangle and P any point on BC. if ` vec P Q` is the sum of ` vec A P` + ` vec P B` +` vec P C` , show that ABPQ is a parallelogram and Q , therefore , is a fixed point.

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To show that \( ABPQ \) is a parallelogram and that \( Q \) is a fixed point, we can follow these steps: ### Step 1: Define the vectors Let \( \vec{A} \), \( \vec{B} \), and \( \vec{C} \) be the position vectors of points \( A \), \( B \), and \( C \) respectively. Let \( \vec{P} \) be the position vector of point \( P \) on line segment \( BC \). ### Step 2: Express the vector \( \vec{PQ} \) According to the problem, we have: \[ ...
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