In [quadrilateral](https://artofproblemsolving.com/wiki/index.php/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 $m + n$.
Extend $\overline{AD}$ and $\overline{BC}$ to meet at $E$. Then, since $\angle BAD = \angle ADC$ and $\angle ABD = \angle DCE$, we know that $\triangle ABD \sim \triangle DCE$. Hence $\angle ADB = \angle DEC$, and $\triangle BDE$ is [isosceles](https://artofproblemsolving.com/wiki/index.php/Isosceles_triangle). Then $BD = BE = 10$. 


Using the similarity, we have:
\[\frac{AB}{BD} = \frac 8{10} = \frac{CD}{CE} = \frac{CD}{16} \Longrightarrow CD = \frac{64}5\]
The answer is $m+n = \boxed{069}$.

Extension: To Find $AD$, use Law of Cosines on $\triangle BCD$ to get $\cos(\angle BCD)=\frac{13}{20}$
Then since $\angle BCD=\angle ABD$ use Law of Cosines on $\triangle ABD$ to find $AD=2\sqrt{15}$