If D, E, F are the midpoints of the sides BC, | vedclass

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(A) Let the position vectors of vertices $A, B, C$ be $\vec{a}, \vec{b}, \vec{c}$ respectively.Since $D, E, F$ are the midpoints of $BC, CA, AB$,their

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$A$ zero vector

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(A) Let the position vectors of vertices $A, B, C$ be $\vec{a}, \vec{b}, \vec{c}$ respectively.Since $D, E, F$ are the midpoints of $BC, CA, AB$,their position vectors are:$\vec{d} = \frac{\vec{b} + \vec{c}}{2}$,$\vec{e} = \frac{\vec{c} + \vec{a}}{2}$,$\vec{f} = \frac{\vec{a} + \vec{b}}{2}$.Now,we calculate the vectors:$\overrightarrow{AD} = \vec{d} - \vec{a} = \frac{\vec{b} + \vec{c}}{2} - \vec{a} = \frac{\vec{b} + \vec{c} - 2\vec{a}}{2}$$\overrightarrow{BE} = \vec{e} - \vec{b} = \frac{\vec{c} + \vec{a}}{2} - \vec{b} = \frac{\vec{c} + \vec{a} - 2\vec{b}}{2}$$\overrightarrow{CF} = \vec{f} - \vec{c} = \frac{\vec{a} + \vec{b}}{2} - \vec{c} = \frac{\vec{a} + \vec{b} - 2\vec{c}}{2}$Adding these vectors:$\overrightarrow{AD} + \overrightarrow{BE} + \overrightarrow{CF} = \frac{(\vec{b} + \vec{c} - 2\vec{a}) + (\vec{c} + \vec{a} - 2\vec{b}) + (\vec{a} + \vec{b} - 2\vec{c})}{2}$$= \frac{(\vec{a} - 2\vec{a} + \vec{a}) + (\vec{b} - 2\vec{b} + \vec{b}) + (\vec{c} - 2\vec{c} + \vec{c})}{2} = \frac{0}{2} = \vec{0}$.Thus,the sum is a zero vector.

If $D, E, F$ are the midpoints of the sides $BC, CA$ and $AB$ of the triangle $ABC$,then $\overrightarrow{AD} + \overrightarrow{BE} + \overrightarrow{CF}$ is:

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