14. Right triangular prism $A B C-A_{1} B_{1} C_{1}$, the base $\triangle A B C$ is an equilateral triangle, $P$ and $E$ are moving points on $B B_{1}$ and $C C_{1}$ (including endpoints), $D$ is the midpoint of side $B C$, and $P D \perp P E$. Then the angle between line $A P$ and $P E$ is $\qquad$
14. $90^{\circ}$.

Since plane $ABC \perp$ plane $BCC_{1}B_{1}, AD \perp BC$, therefore, $AD \perp$ plane $BCC_{1}B_{1}$. Hence $AD \perp PE$. Also, $PE \perp PD, PE \perp$ plane $APD$, then $PE \perp PA$, which means the angle is $90^{\circ}$.