Question

lesson 3 practice problems

1. this diagram is a straightedge and compass construction. a is the center of one circle, and b is the center of the other. select all the true statements.

a. line cd is perpendicular to segment ab
b. point m is the midpoint of segment ab
c. the length ab is the equal to the length cd.
d. segment am is perpendicular to segment bm
e. cb + bd > cd

1. 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 e closer to point a, closer to point b, or the same distance between the points? explain how you know.

Explanation:

Step1: Analyze perpendicular - bisector property

In a straight - edge and compass construction where two circles with centers \(A\) and \(B\) intersect and line \(CD\) is drawn through the intersection points, by the properties of circle - based construction, line \(CD\) is the perpendicular bisector of segment \(AB\). So, line \(CD\) is perpendicular to segment \(AB\) and point \(M\) is the mid - point of segment \(AB\).

Step2: Analyze lengths

There is no reason to assume that \(AB = CD\). Also, \(AM\) and \(BM\) are collinear (since \(M\) is on \(AB\)), so they are not perpendicular.

Step3: Apply triangle inequality

In \(\triangle CBD\), by the triangle inequality theorem, the sum of the lengths of any two sides of a triangle is greater than the length of the third side. So, \(CB + BD>CD\).

Step4: Analyze point \(E\) in second problem

Since the set of points equidistant from \(A\) and \(B\) is the perpendicular bisector of \(AB\) and point \(E\) is not on the perpendicular bisector \(CD\) of \(AB\), we can use the distance formula or the property of perpendicular bisectors. Let's consider right - triangles formed. If we draw lines from \(E\) to \(A\) and \(E\) to \(B\), we can show that the distance from \(E\) to \(A\) and \(E\) to \(B\) are not equal. Since \(E\) is on one side of the perpendicular bisector \(CD\), assume without loss of generality that \(E\) is on the \(A\) - side of \(CD\). Then, \(EA

Answer:

1. A. Line \(CD\) is perpendicular to segment \(AB\); B. Point \(M\) is the midpoint of segment \(AB\); E. \(CB + BD>CD\)
2. Point \(E\) is closer to point \(A\). Since the set of points equidistant from \(A\) and \(B\) is the perpendicular bisector \(CD\) of \(AB\), and \(E\) is on the \(A\) - side of \(CD\), the distance \(EA