Question

5 from unit 1, lesson 1 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. a ab = bc b ab = bd c ad = 2ac d bc = cd e bd = cd

Explanation:

Step1: Analyze the radii of the circles

Since \(A\) is the center of one circle and \(B\) is on the circle, \(AB\) is the radius of the circle centered at \(A\). Since \(C\) is the center of the other circle and \(B\) and \(D\) are on the circle centered at \(C\), \(BC = CD\) (radii of the same circle). Also, \(AD=AC + CD\). If we assume the radius of the circle centered at \(A\) is equal to the radius of the circle centered at \(C\) (by construction with compass, same - width used for radii in such straight - edge and compass constructions), \(AB = BC=CD\). And \(AD=AC + CD\), if \(AC = CD\) (radii of the circle centered at \(C\) with \(A\) and \(D\) endpoints where \(C\) is the center and \(AC\) and \(CD\) are radii in terms of length), then \(AD = 2AC\).

Step2: Check each option

• Option A: \(AB

eq BC\) in general. \(AB\) is the radius of the circle centered at \(A\), and \(BC\) is the radius of the circle centered at \(C\). But by construction (assuming same - compass width for both circles' radii in a proper construction), \(AB = BC\) is wrong.

• Option B: \(AB

eq BD\). \(AB\) is the radius of the circle centered at \(A\), and \(BD\) is not related as a radius of the circle centered at \(A\) in a way that \(AB = BD\).

• Option C: \(AD=AC + CD\). Since \(AC\) and \(CD\) are radii of the circle centered at \(C\) (so \(AC = CD\)), \(AD=2AC\).
• Option D: \(BC\) and \(CD\) are radii of the circle centered at \(C\), so \(BC = CD\).
• Option E: \(BD

eq CD\). \(BD\) is not a radius of the circle centered at \(C\) in the sense that \(BD\) is a chord (not a radius) of the circle centered at \(C\) (unless in a special case, but by the general properties of the construction \(BD
eq CD\)).

Answer:

C. \(AD = 2AC\), D. \(BC = CD\)