Label with $G_1$, $G_2$, $G_3$ the centroids of $\triangle{ABD}$, $\triangle{BCD}$, $\triangle{ADE}$, respectively, and let $AC\cap BE=F$. It is well known that the distance from the midpoint of a side to the centroid is half of the distance between the centroid to its opposite vertex. Hence, $G_1G_2 \| AC$ and $G_1G_3 \| BE$. Therefore, $\angle{G_2G_1G_3}=\angle{EFC}=3\angle{ADB}$. Now notice that, if $AB>r$, and since $\frac{AB}{\sin{\angle{ADB}}}=2r$ then $\sin{\angle{ADB}}>\frac{1}{2} \Rightarrow \frac{\angle{G_2G_1G_3}}{3}=\angle{ADB}>30^{\circ}$, and the result follows.