Question

4 from unit 1, lesson 4 a, b, and c are the centers of the three circles. which line segment is congruent to hf? a ab b cd c df d cb

Explanation:

Step1: Analyze circle radii

Since \( A \), \( B \), \( C \) are circle centers, \( CH = CF = \) radius of circle \( C \), and \( AC = BC = CD = \) radius (assuming equal - radius circles, as the diagram suggests congruent circles). Also, \( HF \) is a chord passing through \( C \), so \( HF = 2\times\text{radius of circle }C \) (diameter of circle \( C \)).

Step2: Analyze each option

- Option A: \( AB \) is a segment between centers \( A \) and \( B \), length is less than \( 2\times\text{radius} \) (since \( A \) and \( B \) are centers of their circles, and the circles intersect, so \( AB \) is less than the sum of two radii, but here we see from the diagram \( AB \) is a short segment, not equal to \( HF \)).
- Option B: \( CD \): \( CD \) is a radius? No, wait, looking at the diagram, \( C \), \( A \), \( B \), \( D \) form a rhombus or equilateral triangle - like figure. Wait, actually, \( HF \) is the diameter of the circle centered at \( C \) (since \( H \) and \( F \) are on the circle centered at \( C \), so \( HF = 2r_C \), where \( r_C \) is the radius of circle \( C \)). \( CD \): Let's see, \( A \), \( B \), \( C \) are centers, so \( AC = BC = CD \) (assuming all circles have the same radius). Wait, no, \( HF \) is the diameter of the top - most circle (center \( C \)), so length \( HF = 2r \) (where \( r \) is the radius of each circle, since the circles seem to have the same radius). Now, \( CD \): Let's check the length. The circle centered at \( A \) and \( B \) and \( C \) have the same radius. So \( AC = r \), \( BC = r \), and \( CD \): from \( C \) to \( D \), \( D \) is on both circles centered at \( A \) and \( B \), so \( AD = r \), \( BD = r \), and \( CD \): since \( A \), \( B \), \( D \) form an equilateral triangle (because \( AD = BD = AB \) if radii are equal), and \( C \) is also equidistant from \( A \) and \( B \), so \( CD \) is equal to \( HF \)? Wait, no, let's re - examine. \( HF \) is the diameter of the circle with center \( C \), so \( HF = 2r \) (where \( r \) is the radius of circle \( C \)). Now, \( CD \): the circle with center \( A \) has radius \( r \), so \( AD = r \), and \( AC = r \) (since \( A \) and \( C \) are centers of their circles, and the circles intersect at \( H \) and \( D \)? Wait, maybe a better approach: congruent segments have the same length. \( HF \) is a horizontal segment through \( C \), length equal to the diameter of the circle centered at \( C \). \( CD \): let's see, the circle centered at \( A \) and \( B \) and \( C \) have the same radius. So \( AC = BC = AB = r \) (if it's an equilateral triangle of centers), and \( CD \): from \( C \) to \( D \), \( D \) is the bottom - most intersection. Wait, maybe the key is that \( HF \) and \( CD \) are both diameters? No, wait, \( HF \) is the diameter of the top circle (center \( C \)), and \( CD \) is the vertical segment from \( C \) to \( D \). Wait, no, let's look at the options again. Wait, the correct approach: in a circle, the radius is the distance from center to a point on the circle. So for circle \( C \), \( CH = CF = r \), so \( HF = CH + CF = 2r \) (since \( H \) and \( F \) are on the circle centered at \( C \), and \( C \) is between \( H \) and \( F \)). Now, for the circle centered at \( A \), \( AD = r \), and for circle centered at \( B \), \( BD = r \), and for circle centered at \( C \), \( CD = r + r = 2r \)? Wait, no, \( C \) to \( D \): if \( A \) and \( B \) and \( C \) are centers of equal - radius circles, then \( AC = r \), \( AD = r \), so triangle \( ACD \) is equilateral? No, maybe \( CD \) is equal to \( HF \) because both are diameters of circles with the same radius. Wait, the correct answer is \( B \) (CD) because \( HF \) is the diameter of the circle with center \( C \), and \( CD \) is the diameter of the circle with center \( A \) or \( B \)? No, wait, let's check the lengths. \( HF \): distance between \( H \) and \( F \), both on circle \( C \), so \( HF = 2\times\text{radius of }C \). \( CD \): distance between \( C \) and \( D \), \( D \) is on circle \( A \) and \( B \), and \( C \) is a center, so if the radius of each circle is \( r \), then \( CD = 2r \) (since \( A \) to \( D \) is \( r \), \( A \) to \( C \) is \( r \), so \( C \) to \( D \) is \( 2r \) in an equilateral triangle - like setup). And \( AB \) is \( r \) (distance between \( A \) and \( B \), centers of two circles), \( DF \): \( D \) to \( F \), \( F \) is on circle \( B \), \( D \) is on circle \( B \), so \( DF \) is a chord of circle \( B \), length less than \( 2r \), \( CB \): \( C \) to \( B \) is \( r \) (distance between centers \( C \) and \( B \)). So \( HF = 2r \) and \( CD = 2r \), so they are congruent.

Answer:

B. CD