2. $A B$ is the diameter of the semicircle $O$, $\odot O_{1}$ is internally tangent to the semicircle at $C_{1}$, and tangent to $A B$ at $D_{1}$; $\odot O_{2}$ is internally tangent to the semicircle at $C_{2}$, and tangent to $A B$ at $D_{2}$ (as shown in Figure 1). The size relationship between $\angle A C_{1} D_{1}$ and $\angle A C_{2} D_{2}$ is ( ).
(A) $\angle A C_{1} D_{1}>\angle A C_{2} D_{2}$
(B) $\angle A C_{1} D_{1}<\angle A C_{2} D_{2}$
(C) $\angle A C_{1} D_{1}=\angle A C_{2} D_{2}$
(D) Cannot be determined
2.C.

The same size, both are $45^{\circ}$. Reason: Obviously, $O 、 O_{1} 、 C_{1}$ are collinear, $O 、 O_{2} 、 C_{2}$ are collinear; also, $O_{1} D_{1} \perp A B, O_{2} D_{2} \perp A B$, so
$$
\begin{aligned}
\angle A C_{1} D_{1} & =\angle A C_{1} O-\angle O C_{1} D_{1} \\
& =\frac{1}{2}\left(\angle C_{1} O B-\angle O O_{1} D_{1}\right), \\
& =\frac{1}{2} \angle O_{1} D_{1} O=\frac{1}{2} \times 90^{\circ}=45^{\circ}, \\
\angle A C_{2} D_{2} & =\angle A C_{2} O+\angle O C_{2} D_{2} \\
& =\frac{1}{2}\left(\angle C_{2} O B+\angle O O_{2} D_{2}\right) \\
& =\frac{1}{2} \times 90^{\circ}=45^{\circ} .
\end{aligned}
$$
$\therefore A C: P_{1}=\angle A C_{2} D_{2}=45^{\circ}$.