Question

unit 1 part 1 lesson 5 practice problems

1. this diagram is a straightedge and compass construction of a line perpendicular to line ab passing through point c. explain why it was helpful to construct points d and a to be the same distance from c.
2. this diagram is a straightedge and compass construction. select all true statements.

a. line ef is the bisector of angle bac.
b. line ef is the perpendicular bisector of segment ba.
c. line ef is the perpendicular bisector of segment ac.
d. line ef is the perpendicular bisector of segment bd.
e. line ef is parallel to line cd.

1. this diagram is a straightedge and compass construction. a is the center of one circle, and b is the center of the other. a rhombus is a quadrilateral with 4 congruent sides. explain why quadrilateral acbd is a rhombus.

Explanation:

Step1: Recall perpendicular - bisector property

In a perpendicular - bisector construction, if we want to construct a perpendicular to a line from a point, making points equidistant from the given point helps in creating congruent triangles. When points \(D\) and \(A\) are the same distance from \(C\), we can use the SSS (Side - Side - Side) congruence criterion for triangles formed during the construction. This allows us to prove that the line constructed is perpendicular to \(AB\).

Step2: Analyze straightedge and compass construction for question 2

For a straightedge and compass construction of a line \(EF\):

• By the properties of such a construction, if \(EF\) is constructed as shown, it is the perpendicular bisector of segment \(BD\). We know that in the construction, the intersection of the arcs centered at points related to \(B\) and \(D\) gives us the perpendicular - bisector. There is no indication that \(EF\) bisects \(\angle BAC\), is the perpendicular bisector of \(BA\) or \(AC\), or is parallel to \(CD\).

Step3: Recall rhombus definition

A rhombus is defined as a quadrilateral with four congruent sides. In the construction where \(A\) and \(B\) are the centers of two circles and \(C\) and \(D\) are the intersection points of the circles, \(AC = AD=BC = BD\) because they are all radii of the two circles. So, quadrilateral \(ACBD\) is a rhombus.

Answer:

1. Making points \(D\) and \(A\) equidistant from \(C\) helps in using triangle - congruence (SSS) to prove the constructed line is perpendicular to \(AB\).
2. c. Line \(EF\) is the perpendicular bisector of segment \(BD\).
3. Since \(A\) and \(B\) are circle - centers and \(C\) and \(D\) are intersection points, \(AC = AD = BC = BD\) (radii), so \(ACBD\) is a rhombus.