14. Right triangular prism $A B C-A_{1} B_{1} C_{1}$, the base $\triangle A B C$ is an equilateral triangle, $\mathrm{P}, \mathrm{E}$ are moving points on $B B_{1}, C C_{1}$ (including endpoints), $\mathrm{D}$ is the midpoint of $\mathrm{BC}$, and $P D \perp P E$. Then the angle between the lines $A P, P E$ is
$90^{\circ}$.
Solution: Since plane $\mathrm{ABC} \perp$ plane $B C C_{1} B_{1}, \mathrm{AD} \perp \mathrm{BC}$, therefore $\mathrm{AD} \perp$ plane $B C C_{1} B_{1}$, so $A D \perp P E$, and since $P E \perp P D, P E \subset$ plane $A P D$, therefore $P E \perp P D$. Thus, the angle is $90^{\circ}$.