2. In $\triangle A B C$, $A B=13, A C=12, B C=$ 5 , take a point $P$ on $A C$. If through $P$ four lines can be drawn such that each line intercepts a triangle similar to $\triangle A B C$, then the maximum value of $C P$ is $\qquad$
2. $\frac{25}{12}$.

As shown in Figure 7, through a point $P$ on $AC$, we can always draw $PD \perp AB$, $PE \parallel BC$, $PG \parallel AB$, to get
$$
\text {Rt } \triangle APD \backsim \text{Rt} \triangle AEP
$$
$\backsim \text{Rt} \triangle PGC \backsim \text{Rt} \triangle ABC$.
The fourth line $PF$ should make
$$
\angle CPF = \angle B,
$$

and point $F$ lies on the line segment $BC$.
From $\triangle FPC \backsim \triangle ABC$, we get
$$
\frac{PC}{BC} = \frac{CF}{CA}.
$$

Thus, $PC = \frac{BC}{AC} \cdot CF \leqslant \frac{BC^2}{AC} = \frac{25}{12}$.
When $CF = CB$, $CP$ reaches its maximum value $\frac{25}{12}$.