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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

\(\overset{\longrightarrow}{OA}+\overset{\longrightarrow}{OB}+\overset{\longrightarrow}{OC} \)\(\overset{\longrightarrow}{OD} + \overset{\longrightarrow}{OE}+\overset{\longrightarrow}{OF}. \)

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Let position vectors of the vertices A, B and C of ΔABC with respect to O be \(\vec a,\vec b\)  and \(\vec c\)respectively.

\(\vec{OA}=\vec a,\) \(\vec {OB}=\vec b\)  and  \(\vec {OC}=\vec c\)

Let us also assume the position vectors of the midpoints D, E and F with respect to O are \(\vec d, \vec e\) and \(\vec f\)respectively.

⇒ \(\vec{OD}=\vec d,\vec{OE}=\vec e\)  and  \(\vec OF=\vec f\)

Now, D is the midpoint of side BC.

This means D divides BC in the ratio 1 : 1.

Recall the position vector of point P which divides AB, the line joining points A and B with position vectors \(\vec a\) and  \(\vec b\) respectively, internally in the ratio m : n is

Similarly, for midpoint E and side CA, we get \(\vec c+\vec a=2\vec e\) and for midpoint F and side AB, we get  \(\vec a+\vec b=2\vec f\)

Adding these three equations, we get

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