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If $D, E$ and $F$ are respectively the mid-points of $A B, A C$ and $B C$ in $\triangle A B C$, then $\overrightarrow{\mathbf{B E}}+\overrightarrow{\mathbf{A F}}$ is equal to :
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Verified Answer
The correct answer is:
$\overrightarrow{\mathbf{D C}}$
Let $A=\overrightarrow{\mathbf{a}}, B=\overrightarrow{\mathbf{b}}, C=\overrightarrow{\mathbf{c}}$
Now, $\quad \overrightarrow{\mathbf{A F}}=\frac{1}{2}(\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}})-\overrightarrow{\mathbf{a}}$
$\overrightarrow{\mathbf{B E}}=\frac{1}{2}(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{c}})-\overrightarrow{\mathbf{b}}$
and $\quad \overrightarrow{\mathbf{C D}}=\frac{1}{2}(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}})-\overrightarrow{\mathbf{c}}$
$\therefore \quad \overrightarrow{\mathbf{A F}}+\overrightarrow{\mathbf{B E}}=\frac{1}{2}(\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}})-\overrightarrow{\mathbf{a}}+\frac{1}{2}(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{c}})-\overrightarrow{\mathbf{b}}$
$=-\frac{1}{2} \overrightarrow{\mathbf{b}}-\frac{1}{2} \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{c}}$
$=\overrightarrow{\mathbf{c}}-\frac{1}{2}(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}})=\overrightarrow{\mathbf{D C}}$
Now, $\quad \overrightarrow{\mathbf{A F}}=\frac{1}{2}(\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}})-\overrightarrow{\mathbf{a}}$
$\overrightarrow{\mathbf{B E}}=\frac{1}{2}(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{c}})-\overrightarrow{\mathbf{b}}$
and $\quad \overrightarrow{\mathbf{C D}}=\frac{1}{2}(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}})-\overrightarrow{\mathbf{c}}$
$\therefore \quad \overrightarrow{\mathbf{A F}}+\overrightarrow{\mathbf{B E}}=\frac{1}{2}(\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}})-\overrightarrow{\mathbf{a}}+\frac{1}{2}(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{c}})-\overrightarrow{\mathbf{b}}$
$=-\frac{1}{2} \overrightarrow{\mathbf{b}}-\frac{1}{2} \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{c}}$
$=\overrightarrow{\mathbf{c}}-\frac{1}{2}(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}})=\overrightarrow{\mathbf{D C}}$
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