4. As shown in Figure 3, in the kite $ABCD$, $AB=AD$, $CB=CD$, the angle bisector of $\angle BDC$ intersects $BC$ at point $L$. Given that points $A$, $L$, $C$, and $D$ are concyclic. Then $2 \angle BAD + 3 \angle BCD$ = . $\qquad$
$4.540^{\circ}$.
As shown in Figure 8, connect
$A L$ intersecting $B D$ at point $M$.
Let $A C$ and $B D$ intersect at point
$O$, and construct the symmetric point $L^{\prime}$ of $L$ with respect to $A C$. It is easy to see that $L^{\prime}$
lies on segment $C D$. Connect $A L^{\prime}$ and $M L^{\{\prime}$.
Since $A, L, C, D$ are concyclic, we have
$$
\begin{array}{l}
\angle C D L=\angle L A C=\angle L^{\prime} A C . \\
\text { By } \angle B M L=\angle A M O=90^{\circ}-\angle L A C \\
=90^{\circ}-\angle C D L=90^{\circ}-\frac{1}{2} \angle B D C \\
=90^{\circ}-\frac{1}{2} \angle D B C,
\end{array}
$$

we know
$$
\begin{array}{l}
\angle B M L=\angle B L M \Rightarrow B M=B L . \\
\text { Also, } \angle M A L^{\prime}=2 \angle L A C \\
=2 \angle C D L=\angle M D L^{\prime},
\end{array}
$$

thus $A, M, L^{\prime}, D$ are concyclic.
By $\frac{B M}{B D}=\frac{B L}{B D}=\frac{C L}{C D}=\frac{C L^{\prime}}{C D}$, we know $M L^{\prime} \parallel B C$.
Therefore, $\angle C B D=\angle L^{\prime} M D=\angle L^{\prime} A D$.
Thus, $\angle O A D=\frac{3}{2} \angle O D C$, which means
$$
\angle B A D=\frac{3}{2}\left(180^{\circ}-\angle B C D\right) \text {. }
$$

Hence, $2 \angle B A D+3 \angle B C D=540^{\circ}$.