Question

the diagram was constructed with straightedge and compass tools. points a, b, c, d, and e are all on line segment cd. name a line segment that is half the length of cd and explain how you know.

Explanation:

Step1: Analyze the construction

The diagram is made with straightedge and compass, so the circles have equal radii. Let the radius of each circle be \( r \). The length of \( CD \) is the distance between \( C \) and \( D \), which is the sum of the diameters of two circles? Wait, no. Looking at the points: \( C \) to \( A \) is a radius, \( A \) to \( B \) is a radius, \( B \) to \( E \) is a radius, \( E \) to \( D \) is a radius? Wait, no, actually, the first circle is centered at \( A \) (or \( C \)? Wait, no, the first circle has center, say, \( A \)? Wait, no, the three circles: the left circle has center, maybe \( A \), the middle circle center \( B \), the right circle center \( E \)? Wait, no, the key is that the distance between \( C \) and \( B \): let's see, \( C \) to \( A \) is a radius, \( A \) to \( B \) is a radius, so \( C \) to \( B \) is \( 2r \). Similarly, \( B \) to \( D \) is \( 2r \)? Wait, no, the length of \( CD \): from \( C \) to \( D \), how many radii? Let's count the segments: \( CA \), \( AB \), \( BE \), \( ED \). Wait, no, the circles: the left circle passes through \( C \) and \( B \) (since \( A \) is the center? Wait, no, if we use compass, when constructing, the radius is the same. So the distance from \( C \) to \( A \) is equal to \( A \) to \( B \), \( B \) to \( E \), and \( E \) to \( D \), because each circle has the same radius. So \( CA = AB = BE = ED = r \) (let \( r \) be the radius). Then \( CD = CA + AB + BE + ED = 4r \)? Wait, no, wait the diagram: \( C \), \( A \), \( B \), \( E \), \( D \) are on \( CD \). The left circle: center at \( A \), radius \( AC \) (so \( AC = AB \), since \( B \) is on the circle centered at \( A \)). The middle circle: center at \( B \), radius \( AB \) (so \( AB = BE \), since \( E \) is on the circle centered at \( B \)). The right circle: center at \( E \), radius \( BE \) (so \( BE = ED \), since \( D \) is on the circle centered at \( E \)). Wait, no, maybe the left circle is centered at \( C \) with radius \( CA \), the middle at \( A \) with radius \( AB \), the right at \( B \) with radius \( BE \), and the last at \( E \) with radius \( ED \). But since it's constructed with compass, the radii are equal. So \( CA = AB = BE = ED \). Let's denote each as \( r \). Then \( CD = CA + AB + BE + ED = 4r \). Now, we need a segment half of \( CD \), so \( \frac{CD}{2} = 2r \). Let's see the segments: \( CB = CA + AB = r + r = 2r \), and \( BD = BE + ED = r + r = 2r \). Also, \( AE = AB + BE = r + r = 2r \)? Wait, \( A \) to \( E \): \( A \) to \( B \) is \( r \), \( B \) to \( E \) is \( r \), so \( AE = 2r \). Wait, but \( CD = 4r \), so half of \( CD \) is \( 2r \). So segments like \( CB \), \( BD \), or \( AE \) would be half of \( CD \). Let's check \( CB \): \( C \) to \( B \). Since the circles are constructed with equal radii, \( AC = AB = BE = ED \). So \( CD = AC + AB + BE + ED = 4 \times AC \) (since all are equal). Then \( CB = AC + AB = 2 \times AC \), so \( CB = \frac{CD}{2} \), because \( CD = 4AC \), so \( \frac{CD}{2} = 2AC = CB \). Similarly, \( BD = BE + ED = 2AC = \frac{CD}{2} \), and \( AE = AB + BE = 2AC = \frac{CD}{2} \). So one such segment is \( CB \) (or \( BD \), or \( AE \)). Let's take \( CB \) as an example.

Step2: Explain the length

Since the diagram is constructed with a compass, the radii of the circles are equal. Let the radius (e.g., \( AC \), \( AB \), \( BE \), \( ED \)) be \( r \). Then \( CD = AC + AB + BE + ED = r + r + r + r = 4r \). The length of \( CB \) is \( AC + AB = r + r = 2r \). Since \( 2r = \frac{4r}{2} = \frac{CD}{2} \), \( CB \) is half the length of \( CD \).

Answer:

A line segment that is half the length of \( CD \) is \( \boldsymbol{CB} \) (or \( BD \), \( AE \)). We know this because the diagram is constructed with a compass, so the radii of the circles are equal (e.g., \( AC = AB = BE = ED \)). Let the length of each radius be \( r \). Then \( CD = AC + AB + BE + ED = 4r \), and \( CB = AC + AB = 2r \). Since \( 2r = \frac{4r}{2} = \frac{CD}{2} \), \( CB \) (or \( BD \), \( AE \)) is half the length of \( CD \).