Transitive Property of Equality Variation 2
Angle Symmetry Example 2
Collinear Angles Property 10
Collinear Angles Property 3
Collinear Angles Property 3 B
Angle Symmetry Example
Transitive Property of Equality Variation 1
Right Angles are Equal
Similar Triangles
Similar Triangles Example 2
Collinear Then 180
Converse of the Supplementary Angles Theorem
Right Angle is 90 Degres
Commutative Property Example 2
Commutative Property Variation 1
Substitution 2
Substitution 8
Subtract Both Sides
Subtraction Example 2
Add Number to Both Sides
Add Number to Both Sides 2
Collinear Angles Property C
Similar Triangles Example 3
Similar Distances 2
Multiplication Property 2
Multiplicative Property of Equality Variation 1
Associative Property of Multiplication 2
Transitive Property Application 2
Substitution 10
Reordering Terms Theorem
Multiplicative Identity 3
Division Theorem
Substitution 11
Substitution 14
Multiplication Property 3
Reorder Terms
Reduction Property
Cross Multiply Theorem
Substitution 12
Substitution in Product
Distance Symmetry Example 5
Distance Symmetry Example 2
Equality Example
Distance Algebra Example
Distributive Property 4
Substitution Example 10
Collinear Points Property 2
Pythagorean Theorem
Pythagorean Theorem 2

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))

A B C X

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)
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))
Previous Lesson Next Lesson

Comments

Please log in to add comments