12. As shown in Figure 3, in the right trapezoid $A B C D$, it is known that $A B \perp A D, A D=D C$ $=1, A B=3$, and a moving point
$P$ is inside a circle centered at point $C$ and tangent to the line $B D$. If
$$
\overrightarrow{A P}=\alpha \overrightarrow{A D}+\beta \overrightarrow{A B}(\alpha, \beta \in \mathbf{R}),
$$

then the range of $\alpha+\beta$ is ( ).
(A) $\left(0, \frac{4}{3}\right)$
(B) $\left(0, \frac{5}{3}\right)$
(C) $\left(1, \frac{4}{3}\right)$
(D) $\left(1, \frac{5}{3}\right)$
12. D.
To find the extremum of $\alpha+\beta$, we might as well assume that point $P$ is on $\odot C$.
It is easy to know that the radius of $\odot C$ is $\frac{1}{\sqrt{10}}$. Therefore,
$$
\begin{aligned}
\overrightarrow{A P}= & \left(1+\frac{1}{\sqrt{10}} \sin \gamma\right) \overrightarrow{A D}+ \\
& \frac{1}{3}\left(1+\frac{1}{\sqrt{10}} \cos \gamma\right) \overrightarrow{A B}(\gamma \in[0,2 \pi)) \\
\Rightarrow & \alpha+\beta \\
= & \frac{4}{3}+\frac{1}{3}\left(\frac{1}{\sqrt{10}} \cos \gamma+\frac{3}{\sqrt{10}} \sin \gamma\right) \in\left[1, \frac{5}{3}\right] .
\end{aligned}
$$

Given that point $P$ is inside $\odot C$, we know that $\alpha+\beta \in\left(1, \frac{5}{3}\right)$.