Question

which rigid transformation would map △abc to △abf?
a reflection across the line containing \\(\overline{cb}\\)
a rotation about point b
a rotation about point a
a reflection across the line containing \\(\overline{ba}\\)

Explanation:

Step1: Analyze the diagram

From the diagram, we can see that \( AC = AF \) (marked with single red ticks), \( \angle CAB=\angle FAB \) (marked with the red arc), and \( AB \) is a common side. So, triangle \( ABC \) and triangle \( ABF \) are symmetric with respect to the line \( AB \)? Wait, no, let's check the angles and sides. Wait, the angle at \( A \) is equal, and \( AC = AF \), \( AB \) is common. Also, looking at the segments, \( BC \) and \( BF \) - wait, maybe a reflection over line \( BA \). Let's check the rigid transformations. A reflection over line \( BA \) (the line containing \( BA \)) would map point \( C \) to \( F \) because \( AC = AF \), \( \angle CAB=\angle FAB \), and \( AB \) is the axis of reflection. Let's check the options:

Option 1: reflection across line containing \( \overline{CB} \) - no, because \( CB \) is not the axis of symmetry here.

Option 2: rotation about point \( B \) - rotating about \( B \) would not map \( C \) to \( F \) as \( BC \) and \( BF \) - not sure, but let's see the other options.

Option 3: reflection across line containing \( \overline{BA} \) - since \( AC = AF \), \( \angle CAB=\angle FAB \), and \( AB \) is the line, reflecting over \( BA \) would map \( C \) to \( F \) and \( B \) stays, \( A \) stays. So triangle \( ABC \) maps to \( ABF \).

Option 4: rotation about point \( A \) - rotating about \( A \), but the angle between \( AC \) and \( AF \) is equal, but the sides \( BC \) and \( BF \) - maybe, but the reflection over \( BA \) is more straightforward. Wait, let's re - examine the diagram. The two triangles share the side \( AB \), \( AC = AF \), and the angle between \( AC \) and \( AB \) is equal to the angle between \( AF \) and \( AB \). So the line of symmetry is \( AB \) (line \( BA \)). So a reflection across the line containing \( \overline{BA} \) (i.e., line \( BA \)) would map \( \triangle ABC \) to \( \triangle ABF \).

Step2: Evaluate each option

• Option "a reflection across the line containing \( \overline{CB} \)": If we reflect over line \( CB \), the image of \( A \) would not land on itself or map to a point such that \( \triangle ABC \) maps to \( \triangle ABF \). So this is incorrect.
• Option "a rotation about point \( B \)": Rotating about point \( B \), the distance from \( B \) to \( C \) and \( B \) to \( F \) - we don't know if they are equal (the diagram doesn't mark \( BC = BF \) with ticks), so rotation about \( B \) is not likely.
• Option "a reflection across the line containing \( \overline{BA} \)": Since \( AC = AF \), \( \angle CAB=\angle FAB \), and \( AB \) is the line of reflection, reflecting \( C \) over \( AB \) gives \( F \), and \( B \) and \( A \) stay on the line, so \( \triangle ABC \) maps to \( \triangle ABF \).
• Option "a rotation about point \( A \)": Rotating about \( A \), we would need to rotate by an angle such that \( C \) maps to \( F \). But since \( \angle CAB=\angle FAB \), the angle of rotation would be \( 0 \) or \( 2\angle CAB \), but the reflection is a more direct transformation here as the triangles are symmetric with respect to line \( BA \).

Answer:

a reflection across the line containing \( \overline{BA} \)