If $D, E$ and $F$ are the mid-points of the sides $B C$, $\mathrm{CA}$ and $\mathrm{AB}$ of triangle…
If $D, E$ and $F$ are the mid-points of the sides $B C$, $\mathrm{CA}$ and $\mathrm{AB}$ of triangle $\mathrm{ABC}$ respectively, then $\overline{\mathrm{AD}}+\frac{2}{3} \overline{\mathrm{BE}}+\frac{1}{3} \overline{\mathrm{CF}}=$
$\frac{1}{2} \overline{\mathrm{AB}}$
$\frac{1}{2} \overline{\mathrm{AC}}$
$\frac{1}{2} \overline{\mathrm{BC}}$
$\frac{2}{3} \overline{\mathrm{AC}}$
Solution
Let the position vector of A, B, C, D, E, F be $\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}, \overline{\mathrm{d}}, \overline{\mathrm{e}}, \overline{\mathrm{f}}$ respectively.
$\begin{aligned}
\therefore \quad \overline{\mathrm{d}} & =\frac{\overline{\mathrm{b}}+\overline{\mathrm{c}}}{2}, \overline{\mathrm{e}}=\frac{\overline{\mathrm{c}}+\overline{\mathrm{a}}}{2}, \overline{\mathrm{f}}=\frac{\overline{\mathrm{a}}+\overline{\mathrm{b}}}{2} \\
& \text { Now, } \overline{\mathrm{AD}}+\frac{2}{3} \overline{\mathrm{BE}}+\frac{1}{3} \overline{\mathrm{CF}} \\
& =\overline{\mathrm{d}}-\overline{\mathrm{a}}+\frac{2}{3}(\overline{\mathrm{e}}-\overline{\mathrm{b}})+\frac{1}{3}(\overline{\mathrm{f}}-\overline{\mathrm{c}}) \\
& =\frac{\overline{\mathrm{b}}+\overline{\mathrm{c}}}{2}-\overline{\mathrm{a}}+\frac{2}{3}\left(\frac{\overline{\mathrm{c}}+\overline{\mathrm{a}}}{2}-\overline{\mathrm{b}}\right)+\frac{1}{3}\left(\frac{\overline{\mathrm{a}}+\overline{\mathrm{b}}}{2}-\overline{\mathrm{c}}\right) \\
& =\frac{\overline{\mathrm{b}}+\overline{\mathrm{c}}-2 \overline{\mathrm{a}}}{2}+\frac{\overline{\mathrm{c}}+\overline{\mathrm{a}}-2 \overline{\mathrm{b}}}{3}+\frac{\overline{\mathrm{a}}+\overline{\mathrm{b}}-2 \overline{\mathrm{c}}}{6} \\
& =\frac{3 \overline{\mathrm{c}}-3 \overline{\mathrm{a}}}{6} \\
& =\frac{3}{6}(\overline{\mathrm{c}}-\overline{\mathrm{a}}) \\
& =\frac{1}{2} \overline{\mathrm{AC}}
\end{aligned}$