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:

View as a tree | View dependent proofs | Try proving it

Given

1	(distance AB) / (distance CB) = (distance BX) / (distance BA)
2	(distance AC) / (distance BC) = (distance CX) / (distance CA)

Proof Table

#	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) (Cross Multiply Theorem)
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) (Cross Multiply Theorem)
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) (Distance Symmetry Example 2)
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) (Distance Symmetry Example 5)
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)) (Equality Example)

Comments

Please log in to add comments