Metamath Proof Explorer


Theorem pythag

Description: Pythagorean theorem. Given three distinct points A, B, and C that form a right triangle (with the right angle at C), prove a relationship between their segment lengths. This theorem is expressed using the complex number plane as a plane, where F is the signed angle construct (as used in ang180 ), X is the distance of line segment BC, Y is the distance of line segment AC, Z is the distance of line segment AB (the hypotenuse), and O is the signed right angle m/__ BCA. We use the law of cosines lawcos to prove this, along with simple trigonometry facts like coshalfpi and cosneg . (Contributed by David A. Wheeler, 13-Jun-2015)

Ref Expression
Hypotheses lawcos.1 ⊢ F = x ∈ ℂ ∖ 0 , y ∈ ℂ ∖ 0 ⟼ ℑ ⁡ log ⁡ y x
lawcos.2 ⊢ X = B − C
lawcos.3 ⊢ Y = A − C
lawcos.4 ⊢ Z = A − B
lawcos.5 ⊢ O = B − C F A − C
Assertion pythag ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ C ∈ ℂ ∧ A ≠ C ∧ B ≠ C ∧ O ∈ π 2 − π 2 → Z 2 = X 2 + Y 2

Proof

Step Hyp Ref Expression
1 lawcos.1 ⊢ F = x ∈ ℂ ∖ 0 , y ∈ ℂ ∖ 0 ⟼ ℑ ⁡ log ⁡ y x
2 lawcos.2 ⊢ X = B − C
3 lawcos.3 ⊢ Y = A − C
4 lawcos.4 ⊢ Z = A − B
5 lawcos.5 ⊢ O = B − C F A − C
6 1 2 3 4 5 lawcos ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ C ∈ ℂ ∧ A ≠ C ∧ B ≠ C → Z 2 = X 2 + Y 2 - 2 ⁢ X ⁢ Y ⁢ cos ⁡ O
7 6 3adant3 ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ C ∈ ℂ ∧ A ≠ C ∧ B ≠ C ∧ O ∈ π 2 − π 2 → Z 2 = X 2 + Y 2 - 2 ⁢ X ⁢ Y ⁢ cos ⁡ O
8 elpri ⊢ O ∈ π 2 − π 2 → O = π 2 ∨ O = − π 2
9 fveq2 ⊢ O = π 2 → cos ⁡ O = cos ⁡ π 2
10 coshalfpi ⊢ cos ⁡ π 2 = 0
11 9 10 eqtrdi ⊢ O = π 2 → cos ⁡ O = 0
12 fveq2 ⊢ O = − π 2 → cos ⁡ O = cos ⁡ − π 2
13 cosneghalfpi ⊢ cos ⁡ − π 2 = 0
14 12 13 eqtrdi ⊢ O = − π 2 → cos ⁡ O = 0
15 11 14 jaoi ⊢ O = π 2 ∨ O = − π 2 → cos ⁡ O = 0
16 8 15 syl ⊢ O ∈ π 2 − π 2 → cos ⁡ O = 0
17 16 3ad2ant3 ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ C ∈ ℂ ∧ A ≠ C ∧ B ≠ C ∧ O ∈ π 2 − π 2 → cos ⁡ O = 0
18 17 oveq2d ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ C ∈ ℂ ∧ A ≠ C ∧ B ≠ C ∧ O ∈ π 2 − π 2 → X ⁢ Y ⁢ cos ⁡ O = X ⁢ Y ⋅ 0
19 subcl ⊢ B ∈ ℂ ∧ C ∈ ℂ → B − C ∈ ℂ
20 19 3adant1 ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ C ∈ ℂ → B − C ∈ ℂ
21 20 3ad2ant1 ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ C ∈ ℂ ∧ A ≠ C ∧ B ≠ C ∧ O ∈ π 2 − π 2 → B − C ∈ ℂ
22 21 abscld ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ C ∈ ℂ ∧ A ≠ C ∧ B ≠ C ∧ O ∈ π 2 − π 2 → B − C ∈ ℝ
23 22 recnd ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ C ∈ ℂ ∧ A ≠ C ∧ B ≠ C ∧ O ∈ π 2 − π 2 → B − C ∈ ℂ
24 2 23 eqeltrid ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ C ∈ ℂ ∧ A ≠ C ∧ B ≠ C ∧ O ∈ π 2 − π 2 → X ∈ ℂ
25 subcl ⊢ A ∈ ℂ ∧ C ∈ ℂ → A − C ∈ ℂ
26 25 3adant2 ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ C ∈ ℂ → A − C ∈ ℂ
27 26 3ad2ant1 ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ C ∈ ℂ ∧ A ≠ C ∧ B ≠ C ∧ O ∈ π 2 − π 2 → A − C ∈ ℂ
28 27 abscld ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ C ∈ ℂ ∧ A ≠ C ∧ B ≠ C ∧ O ∈ π 2 − π 2 → A − C ∈ ℝ
29 28 recnd ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ C ∈ ℂ ∧ A ≠ C ∧ B ≠ C ∧ O ∈ π 2 − π 2 → A − C ∈ ℂ
30 3 29 eqeltrid ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ C ∈ ℂ ∧ A ≠ C ∧ B ≠ C ∧ O ∈ π 2 − π 2 → Y ∈ ℂ
31 24 30 mulcld ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ C ∈ ℂ ∧ A ≠ C ∧ B ≠ C ∧ O ∈ π 2 − π 2 → X ⁢ Y ∈ ℂ
32 31 mul01d ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ C ∈ ℂ ∧ A ≠ C ∧ B ≠ C ∧ O ∈ π 2 − π 2 → X ⁢ Y ⋅ 0 = 0
33 18 32 eqtrd ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ C ∈ ℂ ∧ A ≠ C ∧ B ≠ C ∧ O ∈ π 2 − π 2 → X ⁢ Y ⁢ cos ⁡ O = 0
34 33 oveq2d ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ C ∈ ℂ ∧ A ≠ C ∧ B ≠ C ∧ O ∈ π 2 − π 2 → 2 ⁢ X ⁢ Y ⁢ cos ⁡ O = 2 ⋅ 0
35 2t0e0 ⊢ 2 ⋅ 0 = 0
36 34 35 eqtrdi ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ C ∈ ℂ ∧ A ≠ C ∧ B ≠ C ∧ O ∈ π 2 − π 2 → 2 ⁢ X ⁢ Y ⁢ cos ⁡ O = 0
37 36 oveq2d ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ C ∈ ℂ ∧ A ≠ C ∧ B ≠ C ∧ O ∈ π 2 − π 2 → X 2 + Y 2 - 2 ⁢ X ⁢ Y ⁢ cos ⁡ O = X 2 + Y 2 - 0
38 24 sqcld ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ C ∈ ℂ ∧ A ≠ C ∧ B ≠ C ∧ O ∈ π 2 − π 2 → X 2 ∈ ℂ
39 30 sqcld ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ C ∈ ℂ ∧ A ≠ C ∧ B ≠ C ∧ O ∈ π 2 − π 2 → Y 2 ∈ ℂ
40 38 39 addcld ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ C ∈ ℂ ∧ A ≠ C ∧ B ≠ C ∧ O ∈ π 2 − π 2 → X 2 + Y 2 ∈ ℂ
41 40 subid1d ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ C ∈ ℂ ∧ A ≠ C ∧ B ≠ C ∧ O ∈ π 2 − π 2 → X 2 + Y 2 - 0 = X 2 + Y 2
42 7 37 41 3eqtrd ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ C ∈ ℂ ∧ A ≠ C ∧ B ≠ C ∧ O ∈ π 2 − π 2 → Z 2 = X 2 + Y 2