15. Suppose that in Fig. 2, the length of side of square $A B C D$ is $1, E$ and $F$ are mid-points of $C D$ and $A D$ respectively, $B E$ and $C F$ intersect at a point $P$. Then the length of line segment $C P$ is $\qquad$
15. $\frac{\sqrt{5}}{5}$.

The problem states: As shown in Figure 2, the side length of square $ABCD$ is $1$, and $E$, $F$ are the midpoints of sides $CD$, $AD$ respectively. $BE$ and $CF$ intersect at point $P$. Then the length of segment $CP$ is $\qquad$
From the problem, we know $\triangle CPE \sim \triangle CDF$, so $\frac{CP}{CD} = \frac{CE}{CF}$.
Therefore, $CP = \frac{CD \cdot CE}{CF} = \frac{\sqrt{5}}{5}$.