2. In tetrahedron $ABCD$, $\triangle ABC$ is an equilateral triangle, $AD=BD=2, AD \perp BD, AD \perp CD$, then the distance from $D$ to plane $ABC$ is $\qquad$ .
Given, $A D \perp$ plane $B C D \Rightarrow B D=C D, \triangle A B C$ is an equilateral triangle $\Rightarrow B D \perp C D$, thus the tetrahedron $A B C D$ is a corner of a cube with edge length 2, so the distance from $D$ to the plane $A B C$ is $d=\frac{1}{3} \cdot 2 \sqrt{3}=\frac{2 \sqrt{3}}{3}$.