Using vector method, prove that the medians of a triangle are concurrent.

Answer: All three medians pass through the point \(G\) with position vector \(\frac{\bar{a} + \bar{b} + \bar{c}}{3}\), so they are concurrent.

Let \(\bar{a}, \bar{b}, \bar{c}\) be the position vectors of the vertices \(A, B, C\) of \(\triangle ABC\). Let \(D, E, F\) be the midpoints of \(BC, CA, AB\). Then \(\bar{d} = \frac{\bar{b} + \bar{c}}{2}\), \(\bar{e} = \frac{\bar{c} + \bar{a}}{2}\), \(\bar{f} = \frac{\bar{a} + \bar{b}}{2}\).

Let \(G\) be the point dividing median \(AD\) internally in the ratio 2 : 1. By the section formula,

\(\bar{g} = \frac{2\bar{d} + 1\cdot\bar{a}}{2 + 1} = \frac{(\bar{b} + \bar{c}) + \bar{a}}{3} = \frac{\bar{a} + \bar{b} + \bar{c}}{3}\).

Similarly, the point dividing \(BE\) in the ratio 2 : 1 has position vector \(\frac{2\bar{e} + \bar{b}}{3} = \frac{\bar{a} + \bar{b} + \bar{c}}{3}\), and the point dividing \(CF\) in the ratio 2 : 1 has position vector \(\frac{2\bar{f} + \bar{c}}{3} = \frac{\bar{a} + \bar{b} + \bar{c}}{3}\).

Thus the same point \(G\) lies on all three medians \(AD, BE\) and \(CF\). Hence the medians of a triangle are concurrent (at the centroid \(G\)).