Proof: Distance Algebra Example
Let's prove the following theorem:
if (distance AB) / (distance CB) = (distance BX) / (distance BA) and (distance AC) / (distance BC) = (distance CX) / (distance CA), then ((distance AB) ⋅ (distance AB)) + ((distance AC) ⋅ (distance AC)) = ((distance BC) ⋅ (distance BX)) + ((distance BC) ⋅ (distance CX))
Proof:
Given
1 | (distance AB) / (distance CB) = (distance BX) / (distance BA) |
---|---|
2 | (distance AC) / (distance BC) = (distance CX) / (distance CA) |
# | Claim | Reason |
---|---|---|
1 | (distance AB) ⋅ (distance BA) = (distance CB) ⋅ (distance BX) | if (distance AB) / (distance CB) = (distance BX) / (distance BA), then (distance AB) ⋅ (distance BA) = (distance CB) ⋅ (distance BX) |
2 | (distance AC) ⋅ (distance CA) = (distance BC) ⋅ (distance CX) | if (distance AC) / (distance BC) = (distance CX) / (distance CA), then (distance AC) ⋅ (distance CA) = (distance BC) ⋅ (distance CX) |
3 | (distance AB) ⋅ (distance AB) = (distance BC) ⋅ (distance BX) | if (distance AB) ⋅ (distance BA) = (distance CB) ⋅ (distance BX), then (distance AB) ⋅ (distance AB) = (distance BC) ⋅ (distance BX) |
4 | (distance AC) ⋅ (distance AC) = (distance BC) ⋅ (distance CX) | if (distance AC) ⋅ (distance CA) = (distance BC) ⋅ (distance CX), then (distance AC) ⋅ (distance AC) = (distance BC) ⋅ (distance CX) |
5 | ((distance AB) ⋅ (distance AB)) + ((distance AC) ⋅ (distance AC)) = ((distance BC) ⋅ (distance BX)) + ((distance BC) ⋅ (distance CX)) | if (distance AB) ⋅ (distance AB) = (distance BC) ⋅ (distance BX) and (distance AC) ⋅ (distance AC) = (distance BC) ⋅ (distance CX), then ((distance AB) ⋅ (distance AB)) + ((distance AC) ⋅ (distance AC)) = ((distance BC) ⋅ (distance BX)) + ((distance BC) ⋅ (distance CX)) |
Comments
Please log in to add comments