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Theorem·T28

The Pythagorean theorem

If a triangle has a right angle at C, the squared length of the opposite side equals the sum of the squared lengths of the two sides meeting at C.

Let with (D073). Then with from D072.
In words
For any three points A, B, C in the plane, if there is a right angle at C, then the squared distance from A to B equals the squared distance from A to C plus the squared distance from B to C.
Never needed: F10 · A03 · A04 · A05 · A07 (computed from the citation graph, not asserted).

Proof

  1. 1
    Write , , , and set , , , , so and (D070). The hypothesis unwinds (D073, D071) to
  2. 2
    in terms of . Using associativity, commutativity, and the inverse/identity laws of T26: (the middle step inserts and removes ; the last rewrites using (iv) and (v) applied to T26). Likewise . So (D070).
  3. 3
    Expand the squared distance. By D072: Expanding via distributivity and the sign rules, and using commutativity ( ):
  4. 4
    Use the hypothesis. Negating (H) and using (v) applied to T26 (then commuting): ( is its own inverse since and inverses are unique, L28 (ii)). Regrouping the four negative cross terms from the previous step, via commutativity/associativity:
  5. 5
    Combine. Summing the two expansions and applying the previous step:
  6. 6
    Conclude. By D072 and D070, and (since , ). Reordering the four terms with commutativity/associativity:

Remarks

The oldest and most famous theorem in geometry (Euclid's Elements, Book I, Proposition 47), proved here purely algebraically via the bilinearity of the dot product: , which collapses to exactly when . Working with instead of distance avoids constructing or taking square roots: the theorem already holds over . Example, the familiar 3-4-5 triangle: , , ; then , and . The converse (equal squared distances forces a right angle) and the extension to actual, square-rooted distances over the reals are not developed here.