Example 6 As shown in Figure 1.6.6, E and F are on the sides BC and CD of rectangle ABCD, respectively. If the areas of $\triangle CEF$, $\triangle ABE$, and $\triangle ADF$ are $3$, $4$, and $5$, respectively, find the area of $\triangle AEF$.

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Connect $AC$, let $S_{\triangle AEC}=x, S_{\triangle CAF}=y$. Since $ABCD$ is a rectangle, we have $x+4=y+5$, which means $x=y+1$.
In $\triangle ABC$ and $\triangle AEC$, we have
$$
(x+4): x=BC: EC
$$

In $\triangle ACF$ and $\triangle ECF$, we have

Since $AD=BC$, we have
$$
\begin{array}{c}
y: 3=AD: EC \\
(x+4): x=y: 3
\end{array}
$$

Thus, noting that $x=y+1$, we can solve to get $x=6, y=5$.
Therefore, $S_{\triangle AEF}=x+y-3=8$ is the desired result.
Note: From the solution process, we know that if the condition is changed to parallelogram $ABCD$, the conclusion still holds.