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Question
If O is a point in space, ABC is a triangle and D, E, F are the mid-points of the sides BC, CA and AB respectively of the triangle, prove that \[\vec{OA} + \vec{OB} + \vec{OC} = \vec{OD} + \vec{OE} + \vec{OF}\]
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Solution
Let D, E and F are the midpoints of BC, CA and AB respectively.
Therefore,
\[\frac{\vec{OB} + \vec{OC}}{2} = \vec{OD}\]
\[\vec{OB} + \vec{OC} = 2 \vec{OD} . . . . . \left( 1 \right)\]
Similarly,
\[ \vec{OC} + \vec{OA} = 2 \vec{OE} . . . . . \left( 2 \right)\]
\[ \vec{OA} + \vec{OB} = 2 \vec{OF} . . . . . \left( 3 \right)\]
Adding (1), (2) and (3). We get,
\[2 ( \vec{OA} + \vec{OB} + \vec{OC} ) = 2 ( \vec{OD} + \vec{OE} + \vec{OF} ) . \]
\[ \Rightarrow \vec{OA} + \vec{OB} + \vec{OC} = \vec{OD} + \vec{OE} + \vec{OF} .\]
Hence Proved.
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