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))

Start from the conclusion and work back up the proof. Click the arrow to show the parents.

• ((distance AB) ⋅ (distance AB)) + ((distance AC) ⋅ (distance AC)) = ((distance BC) ⋅ (distance BX)) + ((distance BC) ⋅ (distance CX)),

  if the following are true:

  • a = b
  • c = d

  then a + c = b + d

  • (distance AB) ⋅ (distance AB) = (distance BC) ⋅ (distance BX), if x ⋅ (distance AB) = (distance CD) ⋅ y, then x ⋅ (distance BA) = (distance DC) ⋅ y
    • (distance AB) ⋅ (distance BA) = (distance CB) ⋅ (distance BX),

      if a / b = c / d, then a ⋅ d = b ⋅ c

      • (distance AB) / (distance CB) = (distance BX) / (distance BA)
  • (distance AC) ⋅ (distance AC) = (distance BC) ⋅ (distance CX), if x ⋅ (distance AB) = z, then x ⋅ (distance BA) = z
    • (distance AC) ⋅ (distance CA) = (distance BC) ⋅ (distance CX),

      if a / b = c / d, then a ⋅ d = b ⋅ c

      • (distance AC) / (distance BC) = (distance CX) / (distance CA)