$\begin{array}{l}
\overrightarrow {BI} = \overrightarrow {BC} + \overrightarrow {CI} = \overrightarrow {AD} + \frac{1}{2}\overrightarrow {CD} = \overrightarrow {AD} - \frac{1}{2}\overrightarrow {AB} .\\
\overrightarrow {AG} = \overrightarrow {AB} + \overrightarrow {BG} = \overrightarrow {AB} + \frac{2}{3}\overrightarrow {BM} \,\,\,\left( {M\,\,\,la\,\,\,trung\,\,diem\,\,cua\,\,\,IC} \right)\\
= \overrightarrow {AB} + \frac{2}{3}.\frac{1}{2}\left( {\overrightarrow {BI} + \overrightarrow {BC} } \right)\\
= \overrightarrow {AB} + \frac{1}{3}\overrightarrow {BI} + \frac{1}{3}\overrightarrow {BC} \\
= \overrightarrow {AB} + \frac{1}{3}\overrightarrow {AD} + \frac{1}{3}\left( {\overrightarrow {BA} + \overrightarrow {AD} + \overrightarrow {DI} } \right)\\
= \overrightarrow {AB} + \frac{1}{3}\overrightarrow {AD} - \frac{1}{3}\overrightarrow {AB} + \frac{1}{3}\overrightarrow {AD} + \frac{1}{3}.\frac{1}{2}\overrightarrow {DC} \\
= \frac{2}{3}\overrightarrow {AB} + \frac{2}{3}\overrightarrow {AD} + \frac{1}{6}\overrightarrow {DC} \\
= \frac{2}{3}\overrightarrow {AB} + \frac{2}{3}\overrightarrow {AD} - \frac{1}{6}\overrightarrow {AB} \\
= \frac{1}{2}\overrightarrow {AB} + \frac{2}{3}\overrightarrow {AD} .
\end{array}$
