If \(D, E, F\) are the mid-points of the sides \(BC, CA, AB\) respectively of \(\triangle ABC\), then prove that \(\overline{AD} + \overline{BE} + \overline{CF} = \bar{0}\).

Answer: Proved using position vectors.

Let \(\bar{a}, \bar{b}, \bar{c}\) be the position vectors of \(A, B, C\). By the mid-point formula, the position vectors of \(D, E, F\) are \(\bar{d} = \dfrac{\bar{b} + \bar{c}}{2}\), \(\bar{e} = \dfrac{\bar{c} + \bar{a}}{2}\), \(\bar{f} = \dfrac{\bar{a} + \bar{b}}{2}\).

\(\overline{AD} + \overline{BE} + \overline{CF} = (\bar{d} - \bar{a}) + (\bar{e} - \bar{b}) + (\bar{f} - \bar{c}) = (\bar{d} + \bar{e} + \bar{f}) - (\bar{a} + \bar{b} + \bar{c})\).

Now \(\bar{d} + \bar{e} + \bar{f} = \dfrac{2\bar{a} + 2\bar{b} + 2\bar{c}}{2} = \bar{a} + \bar{b} + \bar{c}\).

\(\therefore \overline{AD} + \overline{BE} + \overline{CF} = \bar{0}\).