14. Given that $A B C-A_{1} B_{1} C_{1}$ is a regular triangular prism, $A B=B C=C A=2, A A_{1}=\sqrt{2}, D$ and $E$ are the midpoints of $A C$ and $B C$ respectively, then the angle formed by $A_{1} D$ and $C_{1} E$ is
14. Take the midpoint $F$ of $A_{1} B_{1}$, then $D E$ is parallel and equal to $\frac{1}{2} A B$ which is parallel and equal to $A_{1} F$, $\therefore E F$ is parallel and equal to $D A_{1}$.
$$
E F=A_{1} D=C_{1} E=\sqrt{1^{2}+(\sqrt{2})^{2}}=\sqrt{3} \text {, and } C_{1} F=\frac{\sqrt{3}}{2} A_{1} B_{1}=\sqrt{3} \text {, }
$$
$\therefore \triangle C_{1} E F_{1}$ is an equilateral triangle, and $\angle F E C_{1}$ is the angle formed by $A_{1} D$ and $C_{1} E$, thus we have.