(i) Two circles $G _ { 1 } , G _ { 2 }$ intersect at points $X , Y$. Choose two other points $A , B$ on $G _ { 1 }$ as shown in the figure. The line segment from $A$ to $X$ is extended to intersect $G _ { 2 }$ at point $L$. The line segment from $L$ to $Y$ is extended to meet $G _ { 1 }$ at point $C$. Likewise the line segment from $B$ to $Y$ is extended to meet $G _ { 2 }$ at point $M$ and the segment from $M$ to $X$ is extended to meet $G _ { 1 }$ at point $D$. Show that $AB$ is parallel to $CD$.
(ii) A triangle $CDE$ is given. A point $A$ is chosen between $D$ and $E$. A point $B$ is chosen between $C$ and $E$ so that $AB$ is parallel to $CD$. Let $F$ denote the point of intersection of segments $AC$ and $BD$. Show that the line joining $E$ and $F$ bisects both segments $AB$ and segment $CD$. (Hint: You may use Ceva's theorem. Alternatively, you may additionally assume that the trapezium $ABCD$ is a cyclic quadrilateral and proceed.)
(iii) Using parts (i) and (ii) describe a procedure to do the following task: given two circles $G _ { 1 }$ and $G _ { 2 }$ intersecting at two points $X$ and $Y$ determine the center of each circle using only a straightedge. Note: Recall that a straightedge is a ruler without any markings. Given two points $A , B$, a straightedge allows one to construct the line segment joining $A , B$. Also, given any two non-parallel segments, we can use a straightedge to find the intersection point of the lines containing the two segments by extending them if necessary.