Example 13 Given in quadrilateral $ABCD$, $\angle ABC=135^{\circ}, \angle BCD=120^{\circ}, AB=\sqrt{6}$, $BC \simeq 5-\sqrt{3}, CD=6$, find the length of $AD$.
Solve: As shown in Figure 1.4.15, draw $D Q \perp B C$ extended to $Q$, draw $A M \perp C B$ extended to $M$, and draw $D N \perp M A$ extended to $N$, then $M N D Q$ is a rectangle.

Given $\angle A B C=135^{\circ}, A B=\sqrt{6}$, we have $M A=M B=\sqrt{3}$. Given $\angle B C D=120^{\circ}, C D=6$, we have $C Q=3, D Q=3 \sqrt{3}$.
Furthermore,
$$
\begin{array}{c}
D N=M Q=M B+B C+C Q=8 \\
A N=M N-A M=Q D-A M=2 \sqrt{3}
\end{array}
$$

Therefore,
$$
A D=\sqrt{A N^{2}+N D^{2}}=2 \sqrt{19}
$$