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:

Step1: Recall properties of reflection

In a reflection, the line of reflection is the perpendicular bisector of the segments connecting corresponding points. Also, reflected figures are congruent.

Step2: Analyze each option

• For a point \(B\) and its image \(B'\) under reflection over line \(EH\), the line \(EH\) bisects the segment \(BB'\), so \(BD = DB'\). This is true.
• There is no information to suggest that \(DF=FG\). This is not related to the properties of reflection between \(\triangle ABC\) and \(\triangle A'B'C'\).
• Since the line of reflection \(EH\) is perpendicular to the segments connecting corresponding - points, \(m\angle EFA = 90^{\circ}\) (assuming \(F\) is on the line of reflection and on the segment connecting a point and its image). This is true.
• The line of reflection \(EH\) is the perpendicular bisector of \(BB'\), \(AA'\), and \(CC'\) as per the definition of reflection. This is true.
• \(\triangle ABC\) and \(\triangle A'B'C'\) are congruent because reflection is a rigid - motion. So the statement \(\triangle ABC\) is not congruent to \(\triangle A'B'C'\) is false.

Answer:

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