Can someone help? It’s about 2 column proofs
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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)