4. Quadrilateral $A B C D$ is a circumscribed quadrilateral of $\odot O$. Then the positional relationship between the incircle $\odot O_{1}$ of $\triangle A B C$ and the incircle $\odot O_{2}$ of $\triangle A C D$ is $\qquad$
4. Externally Tangent.

As shown in Figure 5, let $\odot O_{1}$ and $\odot O_{2}$ be tangent to $AC$ at points $E$ and $F$, respectively. Then,
$$
\begin{array}{l}
A E=\frac{1}{2}(A B+A C-B C), \\
A F=\frac{1}{2}(A D+A C-C D) .
\end{array}
$$

Since quadrilateral $ABCD$ is an ex-tangential quadrilateral of $\odot O$, we have,
$$
A B+C D=A D+B C .
$$

Thus, $A E-A F=\frac{1}{2}(A B+C D-A D-B C)=0$,
which means $A E=A F$.
Therefore, points $E$ and $F$ coincide. Hence, $\odot O_{1}$ and $\odot O_{2}$ are externally tangent.