If $D, E$ and $F$ are respectively the mid-points of $A B, A C$ and $B C$ in $ riangle A B C$, then $\overrightarrow{\mathbf{B E}}+\overrightarrow{\mathbf{A F}}$ is equal to :

Question

If $D, E$ and $F$ are respectively the mid-points of $A B, A C$ and $B C$ in $	riangle A B C$, then $\overrightarrow{\mathbf{B E}}+\overrightarrow{\mathbf{A F}}$ is equal to :

Aicharya · Accepted Answer

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 $\quad \overrightarrow{\mathbf{C D}}=\frac{1}{2}(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}})-\overrightarrow{\mathbf{c}}$ $\therefore \quad \overrightarrow{\mathbf{A F}}+\overrightarrow{\mathbf{B E}}=\frac{1}{2}(\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}})-\overrightarrow{\mathbf{a}}+\frac{1}{2}(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{c}})-\overrightarrow{\mathbf{b}}$ $=-\frac{1}{2} \overrightarrow{\mathbf{b}}-\frac{1}{2} \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{c}}$ $=\overrightarrow{\mathbf{c}}-\frac{1}{2}(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}})=\overrightarrow{\mathbf{D C}}$

If $D, E$ and $F$ are respectively the mid-points of $A B, A C$ and $B C$ in $\triangle A B C$, then…

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