Question

select the counterexample that shows that the conjecture is false. conjecture: a line s divides $overline{mn}$ into two line segments. so, the line s is the segment bisector of $overline{mn}$. if another line also divides $overline{mn}$, then line s is not the only segment bisector of $overline{mn}$. line s must intersect $overline{mn}$ at a right angle in order to be the segment bisector of $overline{mn}$. if line s intersects $overline{mn}$ anywhere besides the mid - point, it is not the segment bisector of $overline{mn}$. if line s intersects $overline{mn}$ at the midpoint, it is the perpendicular bisector of $overline{mn}$.

Explanation:

Step1: Recall definition of segment - bisector

A segment - bisector is a line (or ray or segment) that divides a line segment into two equal parts, i.e., it intersects the line segment at its mid - point.

Step2: Analyze each option

The conjecture says that just because a line divides $\overline{MN}$ into two line segments, it is a segment - bisector. But a line that divides $\overline{MN}$ into two segments is not necessarily a segment - bisector. A line must intersect $\overline{MN}$ at its mid - point to be a segment - bisector. So, if line $s$ intersects $\overline{MN}$ anywhere besides the mid - point, it is not the segment bisector of $\overline{MN}$, which is a counter - example to the conjecture.

Answer:

If line $s$ intersects $\overline{MN}$ anywhere besides the midpoint, it is not the segment bisector of $\overline{MN}$.