OABC is a parallelogram. \(\overrightarrow{OA}\) is represented by the vector \(\mathbf{a}\) and \(\overrightarrow{OC}\) is represented by the vector \(\mathbf{c}\).

1. Find expressions for these vectors, giving the answers in the simplest form: \(\overrightarrow{ON}\), \(\overrightarrow{OM}\), \(\overrightarrow{AN}\) and \(\overrightarrow{AM}\).
2. Show that points A, N and M lie on a straight line.

Point N is \(\frac{2}{3}\) of the distance from O to B.

\(\overrightarrow{ON} = \frac{2}{3} \overrightarrow{OB}\)

\(\overrightarrow{OB} = \mathbf{a} + \mathbf{c}\)

\(\overrightarrow{ON} = \frac{2}{3}(\mathbf{a} + \mathbf{c})\)

\(\overrightarrow{CM} = \frac{1}{2} \overrightarrow{CB}\)

\(\overrightarrow{CM} = \frac{1}{2} \mathbf{a}\)

\(\overrightarrow{OM} = \overrightarrow{OC} + \overrightarrow{CM}\)

\(\overrightarrow{OM} = \mathbf{c} + \frac{1}{2} \mathbf{a}\)

\(\overrightarrow{AN} = \overrightarrow{AO} + \overrightarrow{ON}\)

\(\overrightarrow{AN} = \mathbf{-a} + \frac{2}{3} (\mathbf{a} + \mathbf{c}) = -\frac{1}{3} \mathbf{a} + \frac{2}{3} \mathbf{c}\)

\(\overrightarrow{AM} = \overrightarrow{AO} + \overrightarrow{OM}\)

\(\overrightarrow{AM} = \mathbf{-a} + \frac{1}{2} \mathbf{a} + \mathbf{c} = -\frac{1}{2} \mathbf{a} + \mathbf{c}\)

Show \(\overrightarrow{AM}\) is a multiple of \(\overrightarrow{AN}\).

\(\frac{3}{2} \overrightarrow{AN} = \frac{3}{2} (-\frac{1}{3} \mathbf{a} + \frac{2}{3} \mathbf{c}) = -\frac{1}{2} \mathbf{a} + \mathbf{c} = \overrightarrow{AM}\)

Therefore \(\overrightarrow{AM}\) and \(\overrightarrow{AN}\) are parallel. They also share a common point A so they lie on the same straight line.