Question

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

Explanation:

Step1: Recall properties of circle - circle intersection

In a straight - edge and compass construction of two intersecting circles with centers \(A\) and \(B\), the line joining the intersection points \(C\) and \(D\) is the perpendicular bisector of the line segment joining the centers \(A\) and \(B\). 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 \(M\) lies 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, \(CB + BD>CD\).

Answer:

A. Line \(CD\) is perpendicular to segment \(AB\)
B. Point \(M\) is the midpoint of segment \(AB\)
E. \(CB + BD>CD\)