Let E be the intersection of AC and BD, then E is the midpoint of AC

The medians AM, DE of \(\Delta ADC\) intersect at O, so O is the centroid of \(\Delta ABC\). Then, \(OM=\dfrac{1}{3}AM\)

\(\Delta ODM,\Delta ADM\) have the same altitude drawn from D to AM ; the bases \(OM=\dfrac{1}{3}AM\), so :

\(S_{ODM}=\dfrac{S_{ADM}}{3}=\dfrac{a.\dfrac{a}{2}}{2}:3=\dfrac{a^2}{12}\) (cm²)