Question

△abc was constructed using △abc and line segment eh. for eh to be the line of reflection between △abc and △abc, which statements must be true? select three options. □ bd = db □ df = fg □ m∠efa = 90° □ the line of reflection, eh, is the perpendicular bisector of bb, aa, and cc. □ △abc is not congruent to △abc.

Explanation:

• For \(BD = DB'\):
• In a reflection, the line of reflection is the perpendicular bisector of the segment joining a point and its image. For point \(B\) and its image \(B'\), the line \(EH\) (the line of reflection) bisects \(BB'\). So \(BD = DB'\) (where \(D\) is the intersection of \(BB'\) and \(EH\)).
• For \(m\angle EFA=90^{\circ}\):
• Since \(EH\) is the line of reflection, and \(A\) and \(A'\) are related by reflection over \(EH\). The line segment \(AA'\) is perpendicular to the line of reflection \(EH\). So \(\angle EFA = 90^{\circ}\) (where \(F\) is the intersection of \(AA'\) and \(EH\)).
• For "The line of reflection, \(\overline{EH}\), is the perpendicular bisector of \(\overline{BB'},\overline{AA'}\), and \(\overline{CC'}\)":
• By the definition of a reflection, the line of reflection is the perpendicular bisector of the segments joining pre - image and image points. So \(EH\) is the perpendicular bisector of \(BB'\) (for \(B\) and \(B'\)), \(AA'\) (for \(A\) and \(A'\)) and \(CC'\) (for \(C\) and \(C'\)).
• For \(DF = FG\): There is no information from the reflection property that would make \(DF = FG\). Points \(D\) (from \(BB'\)) and \(F\) (from \(AA'\)) and \(G\) (from \(CC'\)) are not related in a way that \(DF = FG\) based on reflection.
• For "\(\triangle ABC\) is not congruent to \(\triangle A'B'C'\)":
• A reflection is a rigid transformation. Rigid transformations (reflections, rotations, translations) preserve the shape and size of the figure. So \(\triangle ABC\cong\triangle A'B'C'\).

Answer:

• \(BD = DB'\)
• \(m\angle EFA = 90^{\circ}\)
• The line of reflection, \(\overline{EH}\), is the perpendicular bisector of \(\overline{BB'},\overline{AA'}\), and \(\overline{CC'}\)