Answer:
The correct options are;
EFGH has 4 congruent sides
Diagonal FH bisects angles EFG and EHG
Angle FEH is congruent to angle FGH
Step-by-step explanation:
1) Given that for a reflection, we have;
The distance of the reflected preimage from the line of reflection = The distance of the reflected image from the line of reflection
Therefore;
The distance of the point E from the line HF = The distance of the point G from the line HF
Also the reflection of an preimage (x, y) about the x-axis, gives an image (x, -y)
We can show that from the length of a line given by the equation[tex]l = \sqrt{\left (y_{2}-y_{1} \right )^{2}+\left (x_{2}-x_{1} \right )^{2}}[/tex], that the length EH ≅ GH and EF ≅ GF
Therefore since we are given that EH = EF, we have;
EH = GH = GF = EF by the definition of congruency, which gives 4 congruent sides
2) Given that EH = GH = GF = EF and HF = FH by reflective property, we have;
ΔEHF ≅ ΔGHF
∴ ∠GHF ≅ ∠EHF by Congruent Parts of Congruent Triangles are Congruent
Similarly, ∠GFH ≅ ∠EFH
Therefore, ∠GFH = ∠EFH and ∠GHF = ∠EHF
Therefore, diagonal FH bisects angles EFG and EHG
3) Given that ΔEHF ≅ ΔGHF, we have;
Angle FEH is congruent to angle FGH, by Congruent Parts of Congruent Triangles are Congruent
