Proof: Congruent Triangles to Distance 3
Let's prove the following theorem:
if △ABC ≅ △DEF, then distance AC = distance DF
Proof:
Given
| 1 | △ABC ≅ △DEF |
|---|
| # | Claim | Reason |
|---|---|---|
| 1 | distance CA = distance FD | if △ABC ≅ △DEF, then distance CA = distance FD |
| 2 | distance AC = distance DF | if distance CA = distance FD, then distance AC = distance DF |
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