Question

1. triangle fgh is the image of isosceles triangle feh after a reflection across line hf. select all the statements that are a result of corresponding parts of congruent triangles being congruent.

$overline{fe}congoverline{he}$
a. efgh is a rectangle.
b. efgh has 4 congruent sides.
c. diagonal fh bisects angles efg and ehg.
d. diagonal fh is perpendicular to side fe.
e. angle feh is congruent to angle fgh.

Explanation:

Step1: Analyze property of reflection

When a figure is reflected across a line, the corresponding parts of the original figure and its image are congruent.

Step2: Check each option

• Option a: \(EFGH\) is a rhombus (not a rectangle) because \(FE = HE\) (from reflection, corresponding sides are equal) and in a reflection - based congruence of \(\triangle FEH\) and \(\triangle FGH\), all sides \(FE = HE=FG = HG\). So, \(EFGH\) is a rhombus, not a rectangle.
• Option b: Since \(\triangle FEH\cong\triangle FGH\) (by reflection), \(\angle EFG=\angle EHG\). And \(FE = HE = FG=HG\), so \(EFGH\) has 4 congruent sides.
• Option c: Since \(\triangle FEH\cong\triangle FGH\), \(\angle EFH=\angle GFH\). So, diagonal \(FH\) bisects angles \(EFG\) and \(EHG\).
• Option d: There is no information to suggest that diagonal \(FH\) is perpendicular to side \(FE\). In a rhombus \(EFGH\), diagonals are not perpendicular to the sides (they are perpendicular to each other).
• Option e: Since \(\triangle FEH\cong\triangle FGH\) (by reflection), \(\angle FEH=\angle FGH\).

Answer:

b. \(EFGH\) has 4 congruent sides; c. Diagonal \(FH\) bisects angles \(EFG\) and \(EHG\); e. Angle \(FEH\) is congruent to angle \(FGH\)