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 $\qquad$ .
$14.60^{\circ}$.
Take the midpoint $F$ of $A_{1} B_{1}$, then $D E \Perp \frac{1}{2} A B \Perp A_{1} F$. Therefore, $E F \mathbb{\mathbb { I }} D A_{1}$.
$\because E F=A_{1} D=C_{1} E=\sqrt{1^{2}+(\sqrt{2})^{2}}=\sqrt{3}$,
and $C_{1} F=\frac{\sqrt{3}}{2} A_{1} B_{1}=\sqrt{3}$,
$\therefore \triangle C_{1} E F_{1}$ is an equilateral triangle.
Also, $\angle F E C_{1}$ is the angle formed by $A_{1} D$ and $C_{1} E$, thus we get.