a. Proof that the 8-ball at point I is equidistant from holes F and C
Paragraph Proof: To prove that the 8-ball at point I is equidistant from the holes at points F and C, we utilize the definition of a perpendicular bisector. The line segment \( GH \) is the perpendicular bisector of segment \( FC \), meaning it meets segment \( FC \) at its midpoint, J, and forms right angles with \( FC \).
- Definition of Perpendicular Bisector: The definition of a perpendicular bisector states that any point on the bisector is equidistant from the endpoints of the segment it bisects.
- Since point I lies on line segment \( GH \) (the perpendicular bisector of segment \( FC \)), it meets the criteria of the definition.
- Therefore, we conclude that the distance from point I to point F (denoted as \( d(F, I) \)) is equal to the distance from point I to point C (denoted as \( d(C, I) \)): \( d(F, I) = d(C, I) \).
- This confirms that the 8-ball at point I is equidistant from the holes F and C.
b. Name another point in the diagram that would also be equidistant from points F and C
Another point that would be equidistant from points F and C is point J, the intersection point of the segment \( GH \) (the perpendicular bisector) and \( FC \).
Explanation: Since J is defined as the midpoint of segment \( FC \), it is inherently equidistant from both endpoints F and C by the properties of a midpoint. This means \( d(F, J) = d(C, J) \), establishing that not only I, but also J is equidistant from points F and C. Each point on the perpendicular bisector \( GH \), including J, guarantees equality of distances to the endpoints of the segment \( FC \). Thus, whenever a point is on the perpendicular bisector of a segment, it is equidistant from both endpoints of that segment.