2. [4] Let $A, B, C, D, E, F$ be 6 points on a circle in that order. Let $X$ be the intersection of $A D$ and $B E$, $Y$ is the intersection of $A D$ and $C F$, and $Z$ is the intersection of $C F$ and $B E$. $X$ lies on segments $B Z$ and $A Y$ and $Y$ lies on segment $C Z$. Given that $A X=3, B X=2, C Y=4, D Y=10, E Z=16$, and $F Z=12$, find the perimeter of triangle $X Y Z$.
Answer: $\square$
Let $X Y=z, Y Z=x$, and $Z X=y$. By Power of a Point, we have that
$$
3(z+10)=2(y+16), 4(x+12)=10(z+3) \text {, and } 12(x+4)=16(y+2) .
$$

Solving this system gives $X Y=\frac{11}{3}$ and $Y Z=\frac{14}{3}$ and $Z X=\frac{9}{2}$. Therefore, our answer if $X Y+Y Z+$ $Z X=\frac{77}{6}$