Question

Can someone help? It’s about 2 column proofs

Answer 1

Step-by-step explanation:

1. AB = BC (B is the midpoint of AC)

2. DE = EF (E is the midpoint of DF)

3. EB is common

4. ∠ABE = ∠CBE; ∠BED = ∠BEF (EB⊥AC, EB⊥DF)

5. ΔDEB ≅ ΔFEB (RHS)

6. DB = FB (corresponding ∠s of ≅ Δs)

7. ∠EFB = ∠CBF; ∠EDB = ∠ABD (alternate interior angles, AC║DF)

8. ΔABD ≅ ΔCBF (SAS)