jmc

Label point $X$ as shown below, and let $Y$ be the foot of the perpendicular from $X$ to $AD$. Since the hexagon is regular, $\angle DXA=120^{\circ }$ and $\angle AXY=\angle DXY=120^{\circ }/2=60^{\circ }$. Thus, $△AXY$ and $△DXY$ are congruent $30^{\circ }-60^{\circ }-90^{\circ }$ triangles. These triangles are each half an equilateral triangle, so their short leg is half as long as their hypotenuse.

Since the side length of the hexagon is 10, we have $AX=XD=10$. It follows that $XY=AX/2=5$ and $AY=DY=\sqrt{10^2-5^2}=\sqrt{75}=5\sqrt{3}$. (Notice that this value is $\sqrt{3}$ times the length of $XY$, the short leg. In general, the ratio of the sides in a $30^{\circ }-60^{\circ }-90^{\circ }$ is $1:\sqrt{3}:2$, which can be shown by the Pythagorean Theorem.) Then, $DA=2\cdot 5\sqrt{3}=\boxed{10\sqrt{3}}$.