Denote by $V$ the common point of the supporting lines of the lateral edges of the frustum. As planes $(AEF)$ and $(DB'C')$ are parallel, we have $EF \parallel B'C'$ and $DB' \parallel AE$. Thales Theorem gives from $DB' \parallel AE$ and $A'B' \parallel AB$:
$$\frac{VD}{VA} = \frac{VB'}{VE}, \quad \frac{VA'}{VA} = \frac{VB'}{VB}.$$
The last two equalities give
$$\frac{VD}{VA'} = \frac{VB}{VE},$$
so $A'E \parallel DB$. As $EF \parallel B'C' \parallel BC$ and $A'E \parallel DB$ we conclude $(A'EF) \parallel (DBC)$.