Question

the diagram was constructed with straightedge and compass tools.
select all segments that have the same length as segment ac.
a ca
b cb
c ab
d ce
e ad
f ae
g bd
h be
i cd
j ed

Explanation:

Step1: Analyze segment AC

Since the diagram is constructed with straightedge and compass, we assume the circles have equal radii. Let's consider the radii of the circles. Segment \( AC \) is a radius of the left - most circle.

Step2: Analyze each option

• Option A: \( CA \) is the same as \( AC \) (just reversed in direction), so \( CA = AC \).
• Option C: \( AB \): Since the circles are constructed with compass, the distance between the centers (e.g., distance between the centers of the left - most and middle circle, and middle and right - most circle) is equal to the radius. The length of \( AB \) should be equal to the radius (because the circles are constructed such that the radius is consistent). So \( AB=AC \).
• Option H: \( BE \): Similar to \( AB \), \( BE \) is a radius - length segment. Because the circles are constructed with equal radii, \( BE = AC \).
• Option J: \( ED \): \( ED \) is a radius of the right - most circle. Since all circles have the same radius (constructed with compass), \( ED=AC \).
• Option G: \( BD \): \( BD \) is composed of two radius - length segments? Wait, no. Wait, let's re - evaluate. Wait, the left - most circle has center (let's say) at the center of the first circle, with \( AC \) as radius. The middle circle has center at \( B \)? No, wait, the centers of the circles: the first circle (left - most) has center at the mid - point? Wait, no, when we construct circles with compass, if we draw a circle with center at a point and radius \( r \), and then another circle with center at a point \( r \) units away from the first center with the same radius, the intersection points will form equilateral triangles etc. But in terms of radii:
• \( AC \) is a radius. \( AB \) is a radius (since the middle circle is constructed with center at \( A \)? Wait, no, looking at the line \( C - A - B - E - D \). Let's assume that the first circle (left) has center at the center of the first circle, with \( AC \) as radius. The second circle (middle) has center at \( B \)? No, maybe the centers are at \( A \), \( B \), etc. Wait, actually, when we construct circles with compass, if we draw a circle with center at \( A \) and radius \( AC \), and a circle with center at \( B \) and radius \( AC \), then \( AB \) would be equal to \( AC \) if the circles are constructed such that they intersect properly. But more simply, since all the circles are constructed with the same compass setting (radius), the radii of all circles are equal. So:
• \( AC \): radius of left circle.
• \( AB \): radius of middle - left circle (assuming center at \( A \), radius \( AB = AC \)).
• \( BE \): radius of middle - right circle (center at \( B \), radius \( BE=AC \)).
• \( ED \): radius of right circle (center at \( E \), radius \( ED = AC \)).
• \( CA \): same as \( AC \).
• Let's check other options:
• Option B: \( CB=CA + AB=AC + AC = 2AC\) (if \( AB = AC \)), so \( CB

eq AC \).

• Option D: \( CE=CA + AB+BE=AC + AC+AC = 3AC\), so \( CE

eq AC \).

• Option E: \( AD=AB + BE+ED=AC + AC+AC = 3AC\), so \( AD

eq AC \).

• Option F: \( AE=AB + BE=AC + AC = 2AC\), so \( AE

eq AC \).

• Option G: \( BD=BE + ED=AC + AC = 2AC\), so \( BD

eq AC \).

• Option I: \( CD=CA + AB+BE+ED=AC + AC+AC + AC = 4AC\), so \( CD

eq AC \).

Answer:

A, C, H, J