1. As shown in Figure 1, there is a rectangular piece of paper $A B C D, A B=8$, $A D=6$. The paper is folded so that the edge $A D$ lies on the edge $A B$, with the fold line being $A E$. Then, $\triangle A E D$ is folded along $D E$ to the right, and the intersection of $A E$ and $B C$ is point $F$. The area of $\triangle C E F$ is ( ).
(A) 2
(B) 4
(C) 6
(D) 8
- .1.A.

From the folding process, we know $D E=A D=6, \angle D A E=\angle C E F=$ $45^{\circ}$. Therefore, $\triangle C E F$ is an isosceles right triangle, and $E C=8-6$ $=2$. Hence, $S_{\triangle C E F}=2$.