# Construction Techniques 1: Perpendicular Bisectors

6 questions

# Construction Techniques 1: Perpendicular Bisectors

##### Problem 1

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. Write each corresponding letter in the answer box and separate letters with commas.

a) Line $CD$ is perpendicular to segment $AB$ b) Point $M$ is the midpoint of segment $AB$

c) The length $AB$ is equal to the length $CD$ d) Segment $AM$ is perpendicular to segment e) $CB + BD > CD$

##### Problem 2

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.

2) Is point $E$ closer to point $A$, closer to point $B$, or the same distance between the points?

3) Explain how you know.

##### Problem 3

4) Starting with 2 marked points, $A$ and $B$, precisely describe the straightedge and compass moves required to construct the triangle $ABC$ in this diagram.

##### Problem 4

5) This diagram was created by starting with points $C$ and $D$ and using only straightedge and compass to construct the rest. All steps of the construction are visible. Select **all** the steps needed to produce this diagram. Write each corresponding letter in the answer box and separate letters with commas.

a) Construct a circle centered at $A$. b) Construct a circle centered at $C$. c) Construct a circle centered at $D$.

d) Label the intersection points of the circles $A$ and $B$. e) Draw the line through points $C$ and $D$.

f) Draw the line through points $A$ and $B$.

##### Problem 5

6) This diagram was constructed with straightedge and compass tools. $A$ is the center of one circle, and $C$ is the center of the other. Select **all** true statements. Write each corresponding letter in the answer box and separate letters with commas.

a) $AB = BC$ b) $AB = BD$ c) $AD = 2AC$ d) $BC = CD$ e) $BD = CD$