Multiplicative Property of Equality Variation 1
Distance Symmetry Example 4
Transitive Property of Equality Variation 2
Commutative Property Variation 1
Substitution 2
Substitution 6
Distance Symmetry Example 3

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:

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Given
1 (distance BC) ⋅ (distance BC) = ((distance AB) ⋅ (distance AB)) + ((distance AC) ⋅ (distance AC))
Proof Table
# 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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