Question

from unit 1, lesson 3
line segment cd in the following diagram is the perpendicular bisector of line segment ab.
is line segment ab the perpendicular bisector of line segment cd?
explain your reasoning.
5 from unit 1, lesson 3
here are two points in the plane.
a b
a. using only a straightedge, can we find points in the plane that are the same distance from points a and b?
explain your reasoning.
b. using only a compass, can we find points in the plane that are the same distance from points a and b?
explain your reasoning.

Explanation:

Step1: Recall perpendicular - bisector definition

A perpendicular bisector of a line segment divides the segment into two equal parts and is perpendicular to it. Just because $CD$ is the perpendicular bisector of $AB$, it doesn't mean $AB$ is the perpendicular bisector of $CD$ unless additional information is given. In general, we cannot assume reciprocal perpendicular - bisector relationship.

Step2: Analyze straight - edge use

A straightedge can only be used to draw straight lines. To find points equidistant from $A$ and $B$, we need to construct the perpendicular bisector of the line segment $AB$. A straightedge alone cannot construct a perpendicular bisector (we can't measure equal distances or right - angles with just a straightedge).

Step3: Analyze compass use

With a compass, we can draw circles centered at $A$ and $B$ with the same radius. The intersection points of these two circles are equidistant from $A$ and $B$. We can vary the radius of the circles to get multiple such points.

Answer:

1. No. Just because $CD$ is the perpendicular bisector of $AB$ does not imply that $AB$ is the perpendicular bisector of $CD$.
2. a. No. A straightedge can only draw straight lines and cannot be used to construct a perpendicular bisector (which contains points equidistant from $A$ and $B$) without additional tools for measuring right - angles or equal distances.

b. Yes. By drawing circles of the same radius centered at $A$ and $B$, the intersection points of these circles are equidistant from $A$ and $B$.