6. Given that $\odot O_{1}$ and $\odot O_{2}$ are externally tangent at point $P, AB$ is a common external tangent of the two circles, point $A$ is on $\odot O_{1}$, point $B$ is on $\odot O_{2}$, $AC$ is a diameter of $\odot O_{1}$, and $CD$ is a tangent of $\odot O_{2}$, with $D$ being the point of tangency. Then the size relationship between $AC$ and $CD$ is ( ).
(A) $AC > CD$
(B) $AC = CD$
(C) $AC < CD$
(D) Any of the above is possible
6. B.

As shown in Figure 3, connect $A P$, $B P$, and $C P$. Then
$$
\begin{array}{l}
\angle A P C=90^{\circ}, \\
\angle A P B=90^{\circ}. \\
\text { Therefore, } \angle C P B \\
=\angle A P C+ \\
\angle A P B \\
=180^{\circ}.
\end{array}
$$

Thus, points $C$, $B$, and $P$ are collinear.
By the tangent property, $A C \perp A B$.
In the right triangle $\triangle A B C$, by the projection theorem, we have
$$
A C^{2}=C P \cdot C B.
$$

In circle $\odot O_{2}$, by the secant-tangent theorem, we have
$$
C D^{2}=C P \cdot C B.
$$

Therefore, $A C^{2}=C D^{2} \Rightarrow A C=C D$.