Prove as shown in the figure:
12, construct $\square A B P F$, connect $D P$, take the midpoint $M$ of $D P$, then quadrilateral $B C D P$ is a trapezoid. Connect $B_{1} M, E_{1} M$. By the trapezoid midline theorem, we have
$$\begin{array}{l} B_{1} M / / C D / / B P / / A F, \\ M E_{1} / / D E / / F P / / A B, \end{array}$$

and. $\square$
$$\begin{array}{l} B_{1} M=\frac{B P+C D}{2}=\frac{A F+C D}{2}, \\ E_{1} M=\frac{P F+D E}{2}=\frac{A B+D E}{2} . \end{array}$$

Similarly, construct $\square B C D O$, connect $O F$, take the midpoint $N$ of $O F$, connect $A_{1} N, D_{1} N$. By the trapezoid midline theorem, we have
$$\begin{array}{l} A_{1} N / / A F / / B O / / C D, \\ N D_{1} / / E F / / O D / / B C, \end{array}$$

and $A_{1} N=\frac{A F+C D}{2}, D_{1} N=\frac{A B+D E}{2}$.
In $\triangle B_{1} M E_{1}$ and $\triangle A_{1} N D_{1}$,
$$B_{1} M=A_{1} N, E_{1} M=D_{1} N \text {. }$$

Since $A_{1} D_{1}=B_{1} E_{1}$, therefore,
$$\triangle B_{1} M E_{1} \cong \triangle A_{1} N D_{1} \text {. }$$

Thus, $\angle B_{1} M E_{1}=\angle A_{1} N D_{1}$.
Hence $\angle C D E=\angle A F E$.