Proof: Similar Distances
Let's prove the following theorem:
if △ABC ∼ △DEF, then (distance AB) / (distance DE) = (distance AC) / (distance DF)
Proof:
Given
1 | △ABC ∼ △DEF |
---|
# | Claim | Reason |
---|---|---|
1 | (distance AC) / (distance DF) = (distance AB) / (distance DE) | if △ABC ∼ △DEF, then (distance AC) / (distance DF) = (distance AB) / (distance DE) |
2 | (distance AB) / (distance DE) = (distance AC) / (distance DF) | if (distance AC) / (distance DF) = (distance AB) / (distance DE), then (distance AB) / (distance DE) = (distance AC) / (distance DF) |
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