Answer
The proof is in the following form:
Column 1 of proof; Column 2 of proof.
Thus, we have:
1. AB is parallel to DF, and BD is parallel to FG; Given
2. A is congruent to GEF; When parallel lines are cut by a transversal, corresponding angles are congruent.
3. G is congruent to BCA; When parallel lines are cut by a transversal, corresponding angles are congruent.
4. Triangle ABC is congruent to triangle EFG; AA
Work Step by Step
1- Given $\overline{AB} \parallel \overline{DF} $
2- $\angle A = \angle E $ ( corresponding angles of two parallel lines cut by a transversal are congruent )
3- Given $ \overline{BD} \parallel \overline{FG} $
4- $ \angle ABC = \angle EFG $ by the same Reason as step two.
5- $ \triangle ABC \triangle EFG $ are similar by AA
