Question

△abc was constructed using △abc and line segment eh. for \\(\overline{eh}\\) to be the line of reflection between △abc and △abc, which statements must be true? select three options. \\(\square\\) bd = db \\(\square\\) df = fg \\(\square\\) \\(m\angle efa = 90^\circ\\) \\(\square\\) the line of reflection, \\(\overline{eh}\\), is the perpendicular bisector of \\(\overline{bb}\\), \\(\overline{aa}\\), and \\(\overline{cc}\\). \\(\square\\) △abc is not congruent to △abc.

Explanation:

To determine which statements are true for \( \overline{EH} \) being the line of reflection between \( \triangle ABC \) and \( \triangle A'B'C' \), we use the properties of reflections:

Step 1: Analyze \( BD = DB' \)

In a reflection, the distance from a point to the line of reflection is equal to the distance from its image to the line of reflection. So, \( D \) (on \( EH \)) implies \( BD = DB' \). This statement is true.

Step 2: Analyze \( DF = FG \)

There's no reflection property that guarantees \( DF = FG \). These segments aren't related to the reflection of \( \triangle ABC \) to \( \triangle A'B'C' \). This statement is false.

Step 3: Analyze \( m\angle EFA = 90^\circ \)

The line of reflection is perpendicular to the segment joining a point and its image. So, \( EH \perp AA' \), meaning \( \angle EFA = 90^\circ \). This statement is true.

Step 4: Analyze the perpendicular bisector statement

By the definition of a reflection, the line of reflection is the perpendicular bisector of the segments joining each vertex to its image (e.g., \( \overline{BB'} \), \( \overline{AA'} \), \( \overline{CC'} \)). This statement is true.

Step 5: Analyze congruence

Reflections preserve congruence, so \( \triangle ABC \cong \triangle A'B'C' \). 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, \( \overline{EH} \), is the perpendicular bisector of \( \overline{BB'} \), \( \overline{AA'} \), and \( \overline{CC'} \)