$\begin{array}{l}
\overrightarrow {BM} = \overrightarrow {BC} + \overrightarrow {CM} = \overrightarrow {BC} + \dfrac{1}{2}\overrightarrow {CD} \\
= \overrightarrow {BC} + \dfrac{1}{2}\left( {\overrightarrow {CB} + \overrightarrow {BA} + \overrightarrow {AD} } \right)\\
= \overrightarrow {BC} - \dfrac{1}{2}\overrightarrow {BC} - \dfrac{1}{2}\overrightarrow {AB} + \dfrac{1}{2}\overrightarrow {AD} \\
= \dfrac{1}{2}\overrightarrow {BC} - \dfrac{1}{2}\overrightarrow {AB} + \dfrac{1}{2}.3\overrightarrow {BC} \\
= 2\overrightarrow {BC} - \dfrac{1}{2}\overrightarrow {AB} \\
\overrightarrow {AC} = \overrightarrow {AB} + \overrightarrow {BC} \\
\Rightarrow \overrightarrow {BM} .\overrightarrow {AC} = \left( {2\overrightarrow {BC} - \dfrac{1}{2}\overrightarrow {AB} } \right)\left( {\overrightarrow {AB} + \overrightarrow {BC} } \right)\\
= 2\overrightarrow {BC} .\overrightarrow {AB} - \dfrac{1}{2}A{B^2} + 2B{C^2} - \dfrac{1}{2}\overrightarrow {AB} .\overrightarrow {BC} \\
= 2.0 - \dfrac{1}{2}.{\left( {2a} \right)^2} + 2.{a^2} - \dfrac{1}{2}.0\\
= 0\\
\Rightarrow BM \bot AC
\end{array}$
