5. Given that the diameter $A B$ of circle $O$ divides the circle into upper and lower semicircles, and points $C, D$ are on the upper and lower semicircles respectively (neither coinciding with points $A, B$), if $A C=2, A D=1$, then $\overrightarrow{A B} \cdot \overrightarrow{D C}=$ $\qquad$ .
$\begin{array}{l}\text { Solution } \\ \text { As shown in the figure, } \overrightarrow{A B} \cdot \overrightarrow{D C}=\overrightarrow{A B} \cdot(\overrightarrow{A C}-\overrightarrow{A D}) \\ =\overrightarrow{A B} \cdot \overrightarrow{A C}-\overrightarrow{A B} \cdot \overrightarrow{A D} \\ =(\overrightarrow{A C}+\overrightarrow{C B}) \cdot \overrightarrow{A C}-(\overrightarrow{A D}+\overrightarrow{D B}) \cdot \overrightarrow{A D} \\ =\overrightarrow{A C}^{2}-\overrightarrow{A D}^{2}=4-1=3 .\end{array}$