In quadrilateral $ABCD$, $\angle{BAD}\cong\angle{ADC}$ and $\angle{ABD}\cong\angle{BCD}$, $AB=8$, $BD=10$, and $BC=6$. The length $CD$ may be written in the form $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
1. **Extend lines \(BC\) and \(AD\) to meet at point \(E\)**:
   - Since \(\angle{BAD} \cong \angle{ADC}\) and \(\angle{ABD} \cong \angle{BCD}\), triangles \(ABD\) and \(DCE\) are similar by AA similarity criterion.

2. **Establish similarity and corresponding sides**:
   - By similarity, \(\triangle ABD \sim \triangle DCE\).
   - Therefore, the ratios of corresponding sides are equal:
     \[
     \frac{AB}{BD} = \frac{BD}{CD} \quad \text{and} \quad \frac{AB}{BD} = \frac{BD}{DE}
     \]
   - Given \(AB = 8\), \(BD = 10\), and \(BC = 6\), we need to find \(CD\).

3. **Use the similarity ratio**:
   - From the similarity, we have:
     \[
     \frac{AB}{BD} = \frac{BD}{CD}
     \]
   - Substituting the given values:
     \[
     \frac{8}{10} = \frac{10}{CD}
     \]
   - Solving for \(CD\):
     \[
     \frac{8}{10} = \frac{10}{CD} \implies 8 \cdot CD = 10 \cdot 10 \implies 8 \cdot CD = 100 \implies CD = \frac{100}{8} = \frac{25}{2}
     \]

4. **Find \(m\) and \(n\)**:
   - The length \(CD\) is written in the form \(\frac{m}{n}\) where \(m\) and \(n\) are relatively prime.
   - Here, \(CD = \frac{25}{2}\), so \(m = 25\) and \(n = 2\).

5. **Calculate \(m+n\)**:
   - Therefore, \(m+n = 25 + 2 = 27\).

The final answer is \(\boxed{27}\).