2. As shown in Figure 1, in $\triangle A B C$, point $P$ is a point on side $A B$, and it is given that
condition $M: \frac{A C}{A P}=\frac{B C}{C P}$;
condition $N: \triangle A B C \backsim$ $\triangle A C P$.
Then condition $M$ is condition $N$'s ( ).
(A) sufficient but not necessary condition
(B) necessary but not sufficient condition
(C) sufficient and necessary condition
(D) neither sufficient nor necessary condition
2.C.

Obviously, $N \Rightarrow M$. Now we prove: $M \Rightarrow N$.
In $\triangle A B C$ and $\triangle A C P$, using the Law of Sines, we have $\sin \angle A B C=\frac{A C}{B C} \sin A$, $\sin \angle A C P=\frac{A P}{C P} \sin A$.

But from the given condition, we get $\frac{A C}{B C}=\frac{A P}{C P}$, so, $\sin \angle A B C=\sin \angle A C P$.
Since point $P$ is on side $A B$, and
$$
\angle A B C+\angle A C P<\angle A B C+\angle A C B<180^{\circ} \text {, }
$$

then $\angle A C P$ and $\angle A B C$ are not supplementary, thus
$$
\begin{array}{l}
\angle A B C=\angle A C P . \\
\text { Also, } \angle A=\angle A, \text { so, } \triangle A B C \backsim \triangle A C P .
\end{array}
$$