Suppose \$A\$ and \$B\$ are two distinct points in the plane and \$L\$ isthe perpendicular bisector of segment \$overlineAB\$ as pictured below: If \$C\$ is a point on \$L\$, show that \$C\$ is equidistant from \$A\$ and \$B\$,that is show that \$overlineAC\$ and \$overlineBC\$ are congruent.Conversely, show that if \$P\$ is a point which is equidistant from \$A\$and \$B\$, then \$P\$ is on \$L\$.Conclude that the perpendicular bisector of \$overlineAB\$ is exactly the set of pointswhich are equidistant from \$A\$ and \$B\$.

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## IM Commentary

This task is part of a series presenting important foundational geometricresults and constructions which are fundamental for more elaborate arguments.They are presented without a real worldcontext so as to see the important hypotheses and logical steps involved as clearly as possible. Further les-grizzlys-catalans.org, showing how theseresults can be used in context, will also be developed. Teachers should choosean appropriate mixture of abstract and contextual les-grizzlys-catalans.org which best fostersthe development of their students" geometric intuition and understanding.

This task gives the important characterization of the perpendicular bisectorof a line segment as the set of points equidistant from the endpoints of the segment. In the first part of the task, the instructor may need to suggestthat there are two cases to consider:

the case where \$C\$ is on the line segment \$overlineAB\$ (this is a special case andthe students may forget to consider this)the case where \$C\$ is not on the line segment \$overlineAB\$.

The same two cases occur in part (b) for the point \$P\$.

The instructor may also wish to go over the logic of part (b) of the task and make sure that the students understand what needs to be shown and how it is different from part (a). Both (a) and (b) proceed via triangle congruence so the students need to be familiar with and confident in implementing these criteria. Because the task is relatively long and detailed it is recommended mainly for instructional purposes.

This task includes an experimental GeoGebra worksheet, with the intentthat instructors might use it to more interactively demonstrate therelevant content material. The file should be considered a draftversion, and feedback on it in the comment section is highlyencouraged, both in terms of suggestions for improvement and for ideason using it effectively. The file can be run via the free onlineapplication GeoGebra, or runlocally if GeoGebra has been installed on a computer.

This file can be used individually to walk through the given set of problems. It can also be used in a presentation format for an audience. Each step has solutions which can be shown by clicking the check boxes that appear. The buttons are how you switch between steps and at any time you are able to drag the red point on the perpendicular bisector up and down.

## Solution

We consider first the case where \$C\$ is not on segment \$overlineAB\$ as in the picturebelow. The point of intersection of \$L\$ and the segment \$overlineAB\$ is labelled \$O\$: By hypothesis, angles \$COA\$ and \$COB\$ are both right angles and so arecongruent. Side \$overlineCO\$ is congruent to itself and side \$overlineOA\$ is congruent toside \$overlineOB\$ because \$L\$ bisects segment \$overlineAB\$. By SAS we conclude thattriangle \$COA\$ is congruent to triangle \$COB\$. Thus segment \$overlineCA\$ is congruentto segment \$overlineCB\$ and so \$C\$ is equidistant from \$A\$ and \$B\$ as desired.

Next we consider the case where \$C\$ is on segment \$overlineAB\$ as pictured below: This case is simpler than the previous one because we know by hypothesis that\$L\$ bisects segment \$overlineAB\$ and so \$overlineCA\$ is congruent to \$overlineCB\$ and hence\$C\$ is equidistant from \$A\$ and \$B\$ here too.

Here we assume that \$P\$ is equidistant from \$A\$ and \$B\$. There are again twocases to consider. First, \$P\$ could be on segment \$overlineAB\$ and then \$P\$ mustbe the midpoint of segment \$overlineAB\$ and so \$P\$ is on \$L\$. Next suppose \$P\$ isnot on line \$L\$. In this case, we let \$D\$ denote the midpoint of \$overlineAB\$ and \$M\$the line joining \$P\$ and \$D\$ as pictured below: We know that \$overlinePA\$ is congruent to \$overlinePB\$ since \$P\$ is equidistant from \$A\$ and\$B\$ by assumption. We know that \$overlineDA\$ is congruent to \$overlineDB\$ because \$D\$is the midpoint of \$overlineAB\$. Finally \$overlinePD\$ is congruent to \$overlinePD\$. Therefore bySSS, triangle \$PDA\$ is congruent to triangle \$PDB\$. Since angles \$PDA\$ and\$PDB\$ are congruent and add up to \$180\$ degrees, they must both beright angles. Since \$overlineDA\$ is congruent to \$overlineDB\$ we conclude that line \$M\$ isthe perpendicular bisector of segment \$overlineAB\$, that is \$M=L\$.In part (a) we saw that every point on the perpendicular bisector of segment\$overlineAB\$ is equidistant from points \$A\$ and \$B\$. In part (b) we saw that any pointequidistant from points \$A\$ and \$B\$ is on the perpendicular bisector of segment \$overlineAB\$. Therefore the perpendicular bisector of \$overlineAB\$ is made up exactly of thosepoints \$P\$ which are equidistant from \$A\$ and \$B\$.

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Suppose \$A\$ and \$B\$ are two distinct points in the plane and \$L\$ isthe perpendicular bisector of segment \$overlineAB\$ as pictured below: If \$C\$ is a point on \$L\$, show that \$C\$ is equidistant from \$A\$ and \$B\$,that is show that \$overlineAC\$ and \$overlineBC\$ are congruent.Conversely, show that if \$P\$ is a point which is equidistant from \$A\$and \$B\$, then \$P\$ is on \$L\$.Conclude that the perpendicular bisector of \$overlineAB\$ is exactly the set of pointswhich are equidistant from \$A\$ and \$B\$.