Let $\omega$ be a circle, and let $A B C D$ be a quadrilateral inscribed in $\omega$. Suppose that $B D$ and $A C$ intersect at a point $E$. The tangent to $\omega$ at $B$ meets line $A C$ at a point $F$, so that $C$ lies between $E$ and $F$. Given that $A E=6, E C=4, B E=2$, and $B F=12$, find $D A$.
By power of a point, we have $E D \cdot E B=E A \cdot E C$, whence $E D=12$. Additionally, by power of a point, we have $144=F B^{2}=F C \cdot F A=F C(F C+10)$, so $F C=8$. Note that $\angle F B C=$ $\angle F A B$ and $\angle C F B=\angle A F B$, so $\triangle F B C \sim \triangle F A B$. Thus, $A B / B C=F A / F B=18 / 12=3 / 2$, so $A B=3 k$ and $B C=2 k$ for some $k$. Since $\triangle B E C \sim \triangle A E D$, we have $A D / B C=A E / B E=3$, so $A D=3 B C=6 k$. By Stewart's theorem on $\triangle E B F$, we have \[ (4)(8)(12)+(2 k)^{2}(12)=(2)^{2}(8)+(12)^{2}(4) \Longrightarrow 8+k^{2}=8 / 12+12 \] whence $k^{2}=14 / 3$. Thus, \[ D A=6 k=6 \sqrt{14 / 3}=6 \frac{\sqrt{42}}{3}=2 \sqrt{42} \] $\fbox{2\sqrt{42}}$.