We want to find $A D=B C=E F$. So, let $E F$ intersect $B D$ at $O$. It is clear that $\triangle B O E \sim \triangle D O F$. However, we can show by angle chase that $\triangle B X E \sim \triangle D Y F$ : $$\angle B E G=\angle A D G=\angle C B H=\angle D F H$$ This means that $\overline{E F}$ partitions $\overline{B D}$ and $\overline{X Y}$ into the same proportions, i.e. 200 to 80 . Now, let $a=200, b=80, c=9$ to make computation simpler. $O$ is on the radical axis of $\omega_{1}, \omega_{2}$ and its power respect to the two circles can be found to be $$\left(a+\frac{a c}{a+b}\right) \frac{b c}{a+b}=\frac{a b c(a+b+c)}{(a+b)^{2}}$$ However, there is now $x$ for which $O E=a x, O F=b x$ by similarity. This means $x^{2}=\frac{c(a+b+c)}{(a+b)^{2}}$. Notably, we want to find $(a+b) x$, which is just $$\sqrt{c(a+b+c)}=\sqrt{9 \cdot 289}=51$$