Proof: Distance Symmetry Example 3
Let's prove the following theorem:
if (distance BC) ⋅ (distance BC) = ((distance AB) ⋅ (distance AB)) + ((distance AC) ⋅ (distance AC)), then (distance CB) ⋅ (distance CB) = ((distance CA) ⋅ (distance CA)) + ((distance AB) ⋅ (distance AB))
Proof:
Given
1 | (distance BC) ⋅ (distance BC) = ((distance AB) ⋅ (distance AB)) + ((distance AC) ⋅ (distance AC)) |
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# | Claim | Reason |
---|---|---|
1 | (distance BC) ⋅ (distance BC) = (distance CB) ⋅ (distance CB) | (distance BC) ⋅ (distance BC) = (distance CB) ⋅ (distance CB) |
2 | (distance CB) ⋅ (distance CB) = ((distance AB) ⋅ (distance AB)) + ((distance AC) ⋅ (distance AC)) | if (distance BC) ⋅ (distance BC) = (distance CB) ⋅ (distance CB) and (distance BC) ⋅ (distance BC) = ((distance AB) ⋅ (distance AB)) + ((distance AC) ⋅ (distance AC)), then (distance CB) ⋅ (distance CB) = ((distance AB) ⋅ (distance AB)) + ((distance AC) ⋅ (distance AC)) |
3 | ((distance AB) ⋅ (distance AB)) + ((distance AC) ⋅ (distance AC)) = ((distance AC) ⋅ (distance AC)) + ((distance AB) ⋅ (distance AB)) | ((distance AB) ⋅ (distance AB)) + ((distance AC) ⋅ (distance AC)) = ((distance AC) ⋅ (distance AC)) + ((distance AB) ⋅ (distance AB)) |
4 | (distance CB) ⋅ (distance CB) = ((distance AC) ⋅ (distance AC)) + ((distance AB) ⋅ (distance AB)) | if (distance CB) ⋅ (distance CB) = ((distance AB) ⋅ (distance AB)) + ((distance AC) ⋅ (distance AC)) and ((distance AB) ⋅ (distance AB)) + ((distance AC) ⋅ (distance AC)) = ((distance AC) ⋅ (distance AC)) + ((distance AB) ⋅ (distance AB)), then (distance CB) ⋅ (distance CB) = ((distance AC) ⋅ (distance AC)) + ((distance AB) ⋅ (distance AB)) |
5 | (distance AC) ⋅ (distance AC) = (distance CA) ⋅ (distance CA) | (distance AC) ⋅ (distance AC) = (distance CA) ⋅ (distance CA) |
6 | (distance CB) ⋅ (distance CB) = ((distance CA) ⋅ (distance CA)) + ((distance AB) ⋅ (distance AB)) | if (distance CB) ⋅ (distance CB) = ((distance AC) ⋅ (distance AC)) + ((distance AB) ⋅ (distance AB)) and (distance AC) ⋅ (distance AC) = (distance CA) ⋅ (distance CA), then (distance CB) ⋅ (distance CB) = ((distance CA) ⋅ (distance CA)) + ((distance AB) ⋅ (distance AB)) |
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