If D, E and F are respectively the mid-points | 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 mid-points of $AB, AC, BC$ respecti

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$\overrightarrow{DC}$

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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 mid-points of $AB, AC, BC$ respectively, their position vectors are:$\vec{d} = \frac{\vec{a} + \vec{b}}{2}$, $\vec{e} = \frac{\vec{a} + \vec{c}}{2}$, $\vec{f} = \frac{\vec{b} + \vec{c}}{2}$.Now, $\overrightarrow{BE} = \vec{e} - \vec{b} = \frac{\vec{a} + \vec{c}}{2} - \vec{b} = \frac{\vec{a} + \vec{c} - 2\vec{b}}{2}$.And $\overrightarrow{AF} = \vec{f} - \vec{a} = \frac{\vec{b} + \vec{c}}{2} - \vec{a} = \frac{\vec{b} + \vec{c} - 2\vec{a}}{2}$.Adding these two vectors:$\overrightarrow{BE} + \overrightarrow{AF} = \frac{\vec{a} + \vec{c} - 2\vec{b} + \vec{b} + \vec{c} - 2\vec{a}}{2} = \frac{2\vec{c} - \vec{a} - \vec{b}}{2} = \vec{c} - \frac{\vec{a} + \vec{b}}{2}$.Since $\vec{d} = \frac{\vec{a} + \vec{b}}{2}$, we have $\overrightarrow{BE} + \overrightarrow{AF} = \vec{c} - \vec{d} = \overrightarrow{DC}$.

If $D, E$ and $F$ are respectively the mid-points of $AB, AC$ and $BC$ in $\triangle ABC$, then $\overrightarrow{BE} + \overrightarrow{AF}$ is equal to :

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