Example 1 As shown in Figure 4, in tetrahedron $A B C D$, $\triangle A B C$ is an equilateral triangle, $A D=B D=2, A D \perp$ $B D, A D \perp C D$. Then the distance from point $D$

Figure 4 to plane $A B C$ is
[1]
【Analysis】According to the problem, we have
$$
A B=\sqrt{A D^{2}+B D^{2}}=2 \sqrt{2} \text {. }
$$

Then $B C=C A=A B=2 \sqrt{2}$,
$$
\begin{array}{l}
C D=\sqrt{A C^{2}-A D^{2}}=2, \\
A D=B D=C D=2 .
\end{array}
$$

Since $B C^{2}=B D^{2}+C D^{2}$, therefore, $B D \perp C D$.
Taking $D$ as the origin, $\overrightarrow{D A}$ as the $x$-axis, $\overrightarrow{D B}$ as the $y$-axis, and $\overrightarrow{D C}$ as the $z$-axis, we establish a spatial rectangular coordinate system.
It is easy to see that $A(2,0,0), B(0,2,0), C(0,0,2)$.
Thus, the equation of the plane containing $\triangle A B C$ is $x+y+z=2$.
Therefore, the distance from point $D$ to the plane $A B C$ is
$$
\frac{|0+0+0-2|}{\sqrt{1+1+1}}=\frac{2 \sqrt{3}}{3} .
$$