Answer: 5300
Solution 1: Rotate $\triangle A B C$ around $A$ to $\triangle A B^{\prime} C^{\prime}$, such that $B^{\prime}$ is on segment $A F$. Note that as $B D \| E F, A B=A D$. From this, $A B^{\prime}=A B=A D$, and $B^{\prime}=D$. Note that
$$
\angle A D C^{\prime}=\angle A B C=180-\angle A D C,
$$
because $A B C D$ is cyclic. Therefore, $C, D$, and $C^{\prime}$ are collinear. Also, $A C^{\prime}=A C$, and
$$
\angle C A C^{\prime}=\angle D A C+\angle C^{\prime} A D=\angle D A C+\angle C A B=\angle E A F .
$$

Thus, since $A E=A F, \triangle A C C^{\prime} \sim \triangle A E F$. Now, we have
$$
\left[A C C^{\prime}\right]=[A C D]+\left[A D C^{\prime}\right]=[A C D]+[A B C]=[A B C D] .
$$

But, $\left[A C C^{\prime}\right]=\frac{C C^{\prime 2}}{E F^{2}} \cdot[A E F]$, and we know that $C C^{\prime}=C D+D C^{\prime}=4+3=7$. Thus,
$$
\frac{[A B C D]}{[A E F]}=\frac{\left[A C C^{\prime}\right]}{[A E F]}=\frac{7^{2}}{20^{2}}=\frac{49}{400} \text {. }
$$

The answer is $100 \cdot 49+400=5300$.
Solution 2:
Since $B D$ is parallel to $E F$ and $A E=A F$, we have $A B=A D$. Since $A B C D$ is cyclic, $\angle A B C+$ $\angle A D C=180^{\circ}$. Thus we can glue $\triangle A B C$ and $\triangle A D C$ as shown in the diagram above to create a triangle that is similar to $\triangle A E F$ and has the same area as $A B C D$. The base of this triangle has length $B C+C D=3+4=7$, so the desired ratio is
$$
\frac{7^{2}}{20^{2}}=\frac{49}{400} .
$$
5. Let $A E F$ be a triangle with $E F=20$ and $A E=A F=21$. Let $B$ and $D$ be points chosen on segments $A E$ and $A F$, respectively, such that $B D$ is parallel to $E F$. Point $C$ is chosen in the interior of triangle $A E F$ such that $A B C D$ is cyclic. If $B C=3$ and $C D=4$, then the ratio of areas $\frac{[A B C D]}{[A E F]}$ can be written as $\frac{a}{b}$ for relatively prime positive integers $a, b$. Compute $100 a+b$.