Question

1. ab ⊥ cd

in this diagram, line segment cd is the perpendicular bisector of line segment ab. assume the conjecture that the set of points equidistant from a and b is the perpendicular bisector of ab is true. is point m closer to point a, closer to point b, or the same distance from both points? explain how you know.
(from unit 1, lesson 3.)

1. a sheet of paper with points a and b is folded so that a and b match up with each other.

explain why the crease in the sheet of paper is the perpendicular bisector of segment ab. (assume the conjecture that the set of points equidistant from a and b is the perpendicular bisector of segment ab is true.)
(from unit 1, lesson 3.)

1. here is a diagram of a straightedge and compass construction. c is the center of one circle, and b is the center of the other. explain why the length of segment cb is the same as the length of segment cd.

(from unit 1, lesson 1.)

Explanation:

Step 1: Recall the property of perpendicular bisector

The set of points equidistant from \(A\) and \(B\) is the perpendicular bisector of \(AB\). Point \(M\) lies on the perpendicular bisector \(CD\) of \(AB\).

Step 2: Apply the property to point \(M\)

Since \(M\) is on the perpendicular bisector of \(AB\), by the definition of the perpendicular - bisector (a point on the perpendicular bisector of a line segment is equidistant from the endpoints of the line segment), the distance from \(M\) to \(A\) is equal to the distance from \(M\) to \(B\).

for question 7:

Step 1: Analyze the folding action

When we fold the paper so that \(A\) and \(B\) match up, every point on the crease is equidistant from \(A\) and \(B\).

Step 2: Relate to the perpendicular - bisector property

By the conjecture that the set of points equidistant from \(A\) and \(B\) is the perpendicular bisector of \(AB\), the crease in the paper is the perpendicular bisector of segment \(AB\).

for question 8:

Step 1: Consider the construction with circles

In a straight - edge and compass construction, if \(C\) is the center of one circle and \(B\) is the center of the other, and the circles intersect at points \(A\) and \(D\).

Step 2: Use the property of circle radii

\(CA = CD\) because they are radii of the circle with center \(C\), and \(CA=CB\) because the two circles are constructed in such a way that the distance from \(C\) to the intersection point \(A\) (on both circles) is the same as the distance from \(C\) to \(B\) (radius - related property). So, by the transitive property, \(CB = CD\).

Answer:

Point \(M\) is the same distance from both points \(A\) and \(B\).