Let $AEF$ be a triangle with $EF = 20$ and $AE = AF = 21$. Let $B$ and $D$ be points chosen on segments $AE$ and $AF,$ respectively, such that $BD$ is parallel to $EF.$ Point $C$ is chosen in the interior of triangle $AEF$ such that $ABCD$ is cyclic. If $BC = 3$ and $CD = 4,$ then the ratio of areas $\tfrac{[ABCD]}{[AEF]}$ can be written as $\tfrac{a}{b}$ for relatively prime positive integers $a, b$. Compute $100a + b$.
1. **Identify Similar Triangles:**
   Given that $BD \parallel EF$, we can conclude that $\triangle ABD \sim \triangle AEF$ by the AA (Angle-Angle) similarity criterion. This is because $\angle BAD = \angle BAE$ and $\angle ADB = \angle AEF$.

2. **Determine the Ratio of Similar Triangles:**
   Since $\triangle ABD \sim \triangle AEF$, the ratio of their corresponding sides is equal. Let $k$ be the ratio of similarity. Then:
   \[
   \frac{BD}{EF} = \frac{AB}{AE} = \frac{AD}{AF} = k
   \]
   Given $EF = 20$, $AE = AF = 21$, and $BD \parallel EF$, we have:
   \[
   k = \frac{BD}{20}
   \]

3. **Use Cyclic Quadrilateral Properties:**
   Since $ABCD$ is cyclic, we know that opposite angles sum to $180^\circ$. This implies:
   \[
   \angle ACD = \angle ABD \quad \text{and} \quad \angle ACB = \angle ADB
   \]

4. **Rotate and Project Points:**
   Rotate $\triangle ABC$ such that $B$ is projected onto $\overline{AF}$ at point $B'$. Since $AB = AD$, $B' = D$. Project $C$ to point $C'$. Since $C, D, C'$ are collinear, we have:
   \[
   \angle ADC' = \angle ABC = 180^\circ - \angle ADC
   \]

5. **Determine the Lengths and Areas:**
   Since $\triangle ACC' \sim \triangle AEF$, the ratio of their areas is the square of the ratio of their corresponding sides. Given $BC = 3$ and $CD = 4$, we have:
   \[
   C'C = BC + CD = 3 + 4 = 7
   \]
   Therefore, the ratio of the areas is:
   \[
   \left(\frac{C'C}{EF}\right)^2 = \left(\frac{7}{20}\right)^2 = \frac{49}{400}
   \]

6. **Compute the Final Answer:**
   The ratio of the areas $\frac{[ABCD]}{[AEF]}$ is $\frac{49}{400}$. To find $100a + b$ where $\frac{a}{b} = \frac{49}{400}$, we have $a = 49$ and $b = 400$. Thus:
   \[
   100a + b = 100 \times 49 + 400 = 4900 + 400 = 5300
   \]

The final answer is $\boxed{5300}$.