5. As shown in Figure 2, two circles with equal radii intersect at points $B$ and $C$, with centers $O_{1}$ and $O_{2}$ lying on the other circle, respectively. $AB$ is the diameter of $\odot O_{1}$, and $AO_{2}$ intersects $\odot O_{2}$ at points $M$ and $N$, where point $M$ is between $A$ and $O_{2}$. If the extension of $CM$ intersects the extension of $NB$ at point $P$, then $CP: CN=$ $\qquad$
5. $\sqrt{3}: 1$.

Notice that, $\triangle B O_{1} O_{2}$ and $\triangle C O_{1} O_{2}$ are both equilateral triangles.
Thus, $\angle B O_{2} C=120^{\circ}, \angle B N C=60^{\circ}$.
Since $M N$ is the diameter of $\odot O_{2}$,
$\angle M C N=90^{\circ}$.
Therefore, $\triangle P C N$ is a right triangle, and $\angle P=30^{\circ}$.
Hence, $C P: C N=\sqrt{3}: 1$.