[b]Construct a Rhombus[br][/b][br]Knowing what you know about circles and line segments, how might you locate point [math]D[/math] on the ray in the diagram given, so the distance from [math]B[/math] to [math]D[/math] is the same as the distance from [math]B[/math] to [math]A[/math]?
1. Describe how you will locate point [math]D[/math] and how you know [math]BD\cong BA[/math], then construct point [math]D[/math] on the diagram given.
Now that we have three of the four vertices of the rhombus, we need to locate point [math]E[/math], the fourth vertex.[br][br]2. Describe how you will locate point [math]E[/math] and how you will know [math]DE\cong EA\cong AB[/math], then construct point [math]E[/math] on the diagram.
[b]Construct a Perpendicular Bisector of a Segment and a Square (a rhombus with right angles)[br][br][/b]The only difference between constructing a rhombus and constructing a square is that a square contains right angles. Therefore, we need a way to construct perpendicular lines using only a compass and a straight edge.[br][br]We will begin by inventing a way to construct a perpendicular bisector of a line segment.
Describe your strategy for locating points on the perpendicular bisector of [math]RS[/math].
Now that you have created a line perpendicular to [math]RS[/math], we will use the right angle formed to construct a square. [br][br]Label the midpoint of [math]RS[/math] on the diagram as point [math]C[/math]. Using [math]RC[/math] as one side of the square, and the right angle formed by [math]RC[/math] and the perpendicular line drawn through the point [math]C[/math] as the beginning of the square, finish constructing this square on the diagram below. (Hint: Remember that a square is also a rhombus and you have already constructed a rhombus in the first part of the lesson.)
I used circles and lines as construction tools today.[br][br]Take notes as appropriate below.
Circles are useful construction tools because
Congruent circles are useful construction tools because
The congruent circles in the construction of the perpendicular bisector helped me to notice that