To show Mark that the 8-ball at point I is the same distance from the holes at points F and C, we can use perpendicular bisectors. Point M is the midpoint of line FC. Since GH is the perpendicular bisector of FC, it divides FC into two equal parts. This means that IM is also the perpendicular bisector of FC, making point I equidistant from points F and C.
Because point I is on the perpendicular bisector IM, it is equidistant from both F and C. Therefore, the 8-ball at point I is the same distance from the holes at points F and C.
This proves that no matter where Mark hits the 8-ball along the line GH, it will always be equidistant from the holes at points F and C. Mark can aim for either hole with an equal chance of sinking the 8-ball.
rewrite this essay
rewrite this essay and make it more simple To prove to Mark that the 8-ball at point I is equidistant from the holes at points F and C, we can use the properties of perpendicular bisectors.
Let's denote the midpoint of FC¯¯¯¯¯¯¯¯ as point M. Since GH¯¯¯¯¯¯¯¯ is the perpendicular bisector of FC¯¯¯¯¯¯¯¯, it divides FC¯¯¯¯¯¯¯¯ into two equal halves. Therefore, IM¯¯¯¯¯¯¯¯ is also the perpendicular bisector of FC¯¯¯¯¯¯¯¯, and point I is equidistant from points F and C.
Since point I lies on the perpendicular bisector IM¯¯¯¯¯¯¯¯, it is equidistant from both F and C. Thus, the 8-ball at point I is equidistant from the holes at points F and C.
This demonstrates that regardless of where Mark strikes the 8-ball along the perpendicular bisector GH¯¯¯¯¯¯¯¯, it will always be equidistant from the holes at points F and C. Therefore, Mark can aim for either hole with the same probability of sinking the 8-ball.
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