Question

part c
create a point f anywhere on the screen except for on \\(\overleftrightarrow{cd}\\). draw segments \\(\overline{af}\\) and \\(\overline{bf}\\) and find their lengths. what do you notice about the lengths of \\(\overline{af}\\) and \\(\overline{bf}\\).
drag point f around on the screen. where must point f lie so that \\(af = bf\\)?

Explanation:

1. For the first part: When you move point F (not on $\overleftrightarrow{CD}$), the lengths $\overline{AF}$ and $\overline{BF}$ are only equal in specific positions, not everywhere.
2. For the second part: The set of all points equidistant from two fixed points A and B is the perpendicular bisector of the segment connecting A and B. Since $\overleftrightarrow{CD}$ is implied to be the perpendicular bisector of $\overline{AB}$ (from the context of equal length hint at the top), point F must lie on $\overleftrightarrow{CD}$ to have $AF = BF$.

Answer:

1. For most positions of F (not on $\overleftrightarrow{CD}$), the lengths of $\overline{AF}$ and $\overline{BF}$ are not equal; they only match in specific locations.
2. Point F must lie on $\overleftrightarrow{CD}$ (the perpendicular bisector of $\overline{AB}$) to have $AF = BF$.