Example 6. As shown in the figure, in pentagon $A B C D E$, $\angle B=\angle A E D=$ $90^{\circ}, A B=C D=A E=B C+D E=1$. Find the area of this pentagon. (1992, Beijing Junior High School Grade 2 Mathematics Competition)
Analysis We immediately think of connecting $A C, A D$, because there are two right-angled triangles. But we also find that it is not easy to directly calculate the area of each triangle. Given the condition $B C+D E=1$, we wonder if we can join $B C$ to one end of $D E$, making $E F=B C$. Connect $A F$. Then we find $\triangle A B C \cong$ $\triangle A E F$. So $F D=1$, and $=A C \cdot \triangle E=A B, \triangle A D F$ are triangles with bases of 1, each with an area of $\frac{1}{2}$, and $\triangle A C D$ is congruent to $\triangle A D F$, thus the total area is 1.