Question

your answer
i think point e is closer to point a than point b because all points located on line segment cd are equidistant to ae and f is on the a side.

feedback
your reasoning has the right idea but contains some inaccuracies. you correctly identify that point e is closer to point a, but your explanation needs clarification.

since cd is the perpendicular bisector of ab, any point on the line cd would be equidistant from a and b. however, point e appears to be on the same side of the perpendicular bisector as point a, not on the line cd itself.
points that are not on the perpendicular bisector will be closer to whichever endpoint is on their side of the perpendicular bisector.

so e is closer to a because its on the a side of the perpendicular bisector,
not because its on the line cd.

Explanation:

To address the feedback and correct the explanation:

Step 1: Recall the Perpendicular Bisector Theorem

The Perpendicular Bisector Theorem states that any point on the perpendicular bisector of a segment is equidistant from the segment’s endpoints. For segment \( AB \) with perpendicular bisector \( CD \), this means:
If a point \( P \) lies on \( CD \), then \( PA = PB \).

Step 2: Analyze Point \( E \)’s Position

Point \( E \) is not on \( CD \) (the perpendicular bisector) but on the “\( A \)-side” of \( CD \). Points not on the perpendicular bisector are closer to the endpoint on their side of the bisector.

Since \( E \) is on the \( A \)-side of \( CD \) (the perpendicular bisector of \( AB \)), it is closer to \( A \) than to \( B \).

Answer:

Point \( E \) is closer to point \( A \) because \( E \) lies on the \( A \)-side of the perpendicular bisector \( CD \) of \( AB \). Points not on the perpendicular bisector are closer to the endpoint on their side of the bisector, so \( E \) is closer to \( A \) (not because it is on \( CD \), but because it is on \( A \)’s side of \( CD \)).