Proof: Congruent Triangles to Angles 2
Let's prove the following theorem:
if △ABC ≅ △DEF, then m∠EFD = m∠ACB
Proof:
Given
| 1 | △ABC ≅ △DEF |
|---|
| # | Claim | Reason |
|---|---|---|
| 1 | m∠BCA = m∠EFD | if △ABC ≅ △DEF, then m∠BCA = m∠EFD |
| 2 | m∠EFD = m∠BCA | if m∠BCA = m∠EFD, then m∠EFD = m∠BCA |
| 3 | m∠BCA = m∠ACB | m∠BCA = m∠ACB |
| 4 | m∠EFD = m∠ACB | if m∠EFD = m∠BCA and m∠BCA = m∠ACB, then m∠EFD = m∠ACB |
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