Question

Please help me. I need help please.

Answer 1

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$l = \sqrt{\left (y_{2}-y_{1} \right )^{2}+\left (x_{2}-x_{1} \right )^{2}}$, 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