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△abc was constructed using △abc and line segment eh. for eh to be the l…

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 reflection properties

In a reflection, the line of reflection is the perpendicular bisector of the segments connecting corresponding points. So, for points \(B\) and \(B'\), the line \(EH\) (line of reflection) is the perpendicular bisector of \(BB'\), which means \(BD = DB'\).

Step2: Analyze angle - related property

The line of reflection is perpendicular to the segments connecting corresponding points. So, the angle between the line of reflection \(EH\) and the segment \(AA'\) (at the intersection point \(F\)) is \(90^{\circ}\), i.e., \(m\angle EFA=90^{\circ}\).

Step3: Consider congruence and bisector property

A reflection is a rigid transformation, so \(\triangle ABC\cong\triangle A'B'C'\), and the line of reflection \(EH\) is the perpendicular bisector of \(BB'\), \(AA'\) and \(CC'\) as it is the line of reflection. There is no information to suggest \(DF = FG\).

Answer:

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