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 ab ⊥ cd

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:

Problem 1

• Option A: In a straightedge and compass construction with two circles centered at A and B, the line through their intersection points (CD) is the perpendicular bisector of AB, so \( CD \perp AB \). True.
• Option B: Since CD is the perpendicular bisector of AB, M (the intersection of AB and CD) is the midpoint of AB. True.
• Option C: AB is a segment connecting centers, CD is a chord (or the perpendicular bisector segment). Their lengths aren’t necessarily equal. False.
• Option D: AM and BM are colinear (part of AB), so they can’t be perpendicular. False.
• Option E: By the triangle inequality, in \( \triangle CBD \), \( CB + BD > CD \) (sum of two sides > third side). True.

The set of points equidistant from A and B lies on the perpendicular bisector (CD) of AB. Point E is not on CD. So, E is closer to the point (A or B) whose side of CD it lies on. From the diagram, E is on the side of A relative to CD. Thus, E is closer to A (because points on CD are equidistant, and E is off - CD towards A, so its distance to A is less than to B).

Answer:

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

Problem 2