Question

1. triangle fgh is the image of isosceles triangle feh after a reflection across line hf. (overline{eh}congoverline{ef}) select all the statements that are a result of corresponding parts of congruent triangles being congruent.

a. quadrilateral efgh is a rectangle.
b. quadrilateral efgh is a rhombus.
c. diagonal fh bisects angles efg and ehg.
d. diagonal fh is perpendicular to side fe.
e. angle ehf is congruent to angle fgh.
f. angle feh is congruent to angle fgh.

Explanation:

Step1: Analyze the properties of congruent triangles and reflection

Since \(\triangle FGH\) is the image of \(\triangle FEH\) after a reflection across line \(HF\), we have \(\triangle FEH\cong\triangle FGH\). So \(EH = GH\), \(EF=FG\), \(\angle EHF=\angle GHF\), \(\angle EFH=\angle GFH\), \(\angle FEH=\angle FGH\)

Step2: Analyze each option

• Option A:

For a quadrilateral to be a rectangle, all angles must be right - angles. There is no information to suggest that the angles of quadrilateral \(EFGH\) are right - angles. So, quadrilateral \(EFGH\) is not a rectangle.

• Option B:

Since \(EH = GH\), \(EF = FG\) and \(\triangle FEH\cong\triangle FGH\), \(EH=EF\) (given \(\overline{EH}\cong\overline{EF}\)). So \(EH = EF=FG = GH\). A quadrilateral with all sides equal is a rhombus.

• Option C:

Because \(\triangle FEH\cong\triangle FGH\) and reflection across \(HF\), \(\angle EFH=\angle GFH\) and \(\angle EHF=\angle GHF\). So diagonal \(FH\) bisects \(\angle EFG\) and \(\angle EHG\)

• Option D:

There is no information to suggest that \(\angle EFH = 90^{\circ}\). So diagonal \(FH\) is not necessarily perpendicular to side \(FE\)

• Option E:

\(\angle EHF\) and \(\angle FGH\) are not corresponding angles of congruent triangles \(\triangle FEH\) and \(\triangle FGH\)

• Option F:

Since \(\triangle FEH\cong\triangle FGH\), by the property of congruent triangles (CPCTC - Corresponding Parts of Congruent Triangles are Congruent), \(\angle FEH\cong\angle FGH\)

Answer:

B. Quadrilateral \(EFGH\) is a rhombus; C. Diagonal \(FH\) bisects angles \(EFG\) and \(EHG\); F. Angle \(FEH\) is congruent to angle \(FGH\)