6. Let $I$ be the incenter of $\triangle A B C$, $r$ be the radius of its incircle, and $R$ be the radius of its circumcircle. If the extension of $A I$ intersects the circumcircle of $\triangle A B C$ at point $P$, then $I A \cdot I P=(\quad)$.
(A) $2 R_{r}$
(B) $2 R^{2} r$
(C) $2 R r^{2}$
(D) $2 R^{2} r^{2}$
6. A.

As shown in Figure 3, construct the diameter $PQ$ of the circumcircle $\odot O$ of $\triangle ABC$, and connect $BQ$, $BP$. Let $\odot I$ be tangent to $AB$ at point $D$, and connect $IB$, $ID$. It is easy to see that
$$
\begin{array}{l}
\angle PQB = \angle PAB \\
= \angle PAC = \angle PBC.
\end{array}
$$

Therefore, we have
$$
\begin{array}{c}
\text{Rt } \triangle PBQ \sim \text{Rt } \triangle IDA. \\
\text{Thus, } \frac{PQ}{IA} = \frac{PB}{ID}, \text{ i.e., } \\
IA \cdot PB = PQ \cdot ID = 2Rr.
\end{array}
$$

Since $\angle IBC = \angle IBA$, then $\angle PBI = \angle PIB$. Therefore, $PB = IP$. Hence, $IA \cdot IP = 2Rr$.