Answer:
Option B is correct.
Statement: ∠AFE ≅ ∠AFD
Reason: Transitive Property of Equality
Explanation:
Given that:
[tex]\overline{BC} || \overline{DE}[/tex] and [tex]\angle GAC \cong \angle AFD[/tex]
Prove that : [tex]\overleftrightarrow{GH} \perp \overleftrightarrow{DE}[/tex]
Proof:
Given:
[tex]\angle GAC \cong \angle AFD[/tex] ....[1]
By theorem of corresponding angles:
For parallel lines cut by transversal , the corresponding angles are congruent.
then;
[tex]\angle GAC \cong \angle AFE[/tex] ....[2]
Transitive property of equality states that if x = y and y = z then; x = z
Using transitive property of equality in [1] and [2] we have;
[tex]\angle AFE \cong \angle AFD[/tex]
By Linear pair theorem states that if the two pairs are adjacent to each other then they are supplementary.
⇒[tex]\angle AFD[/tex] and [tex]\angle AFE[/tex] are supplementary.
We know that:
If two congruent angles are supplementary, then each angle is of right angle i.,e 90 degree
⇒[tex]m\angle AFD =m\angle AFE = 90^{\circ}[/tex]
By definition of perpendicular line:
If two line meet at a right angle i.,e 90 degree, then those lines are perpendicular.
⇒ [tex]\overleftrightarrow{GH} \perp \overleftrightarrow{DE}[/tex] Hence proved!
