Question

name:
geometry
date
per:
1a) this diagram is a straightedge and compass construction. ( c ) is the center of both circles.
select all statements that must be true by construction. (circle them)
a segments ( ab ) and ( ad ) have the same length.
b segments ( ac ) and ( ad ) have the same length.
c segments ( ac ) and ( cd ) have the same length.
d triangle ( bce ) is isosceles.
e triangle ( cde ) is isosceles.
1b) justify the answers you have chosen above. you must give reason(s) for each selection.

1. construct a perpendicular bisector through segment ( ab ). show constructions clearly. no credit given for estimated drawings. (note: point ( c ) is an arbitrary point for a question below)

( \text{a)} ) construction
( \text{b)} ) mark the resulting point of intersection with the letter ( d )
( \text{c)} ) what is the relation of point ( d )? how do you know? justify using your construction
( \text{d)} ) is point ( c ) closer to point ( a ), closer to point ( b ) or equidistant to them? justify using your construction
3a) construct the following polygon using the instructions below. be sure to label precisely

1. draw 2 points: ( a ) and ( b ).
2. draw a circle centered at ( a ) with radius ( ab ).
3. draw a circle centered at ( b ) with radius ( ab ).
4. label the intersection points of the circles ( c ) and ( d ).
5. draw segments ( ac ), ( bc ), ( ad ), and ( bd ).

3b) what polygon was created? how do you know?

Explanation:

Answer:

1a)
A. Segments AB and AD have the same length
C. Segments AC and CE have the same length
D. Triangle BCD is isosceles
1b)

• A: AB and AD are radii of the larger circle centered at C.
• C: AC and CE are radii of the smaller circle centered at C.
• D: BC = DC (radii of larger circle), so △BCD is isosceles.

2a) [Draw arcs above and below AB from A and B with radius >$\frac{1}{2}$AB, connect arc intersections]
2b) [Mark intersection point as D]
2c) Point D is equidistant from A and B; it lies on the perpendicular bisector of AB.
2d) Point C is closer to [whichever is visually closer]; perpendicular bisector of AB has points equidistant to A and B, C is not on it.
3a) [Follow steps to draw two intersecting circles, label and connect points]
3b) Rhombus; all sides (AB=AC=BC=AD=BD) are equal length.