Metamath Proof Explorer


Theorem perprag

Description: Deduce a right angle from perpendicular lines. (Contributed by Thierry Arnoux, 10-Nov-2019)

Ref Expression
Hypotheses colperpex.p ⊢ P = Base G
colperpex.d ⊢ - ˙ = dist ⁡ G
colperpex.i ⊢ I = Itv ⁡ G
colperpex.l ⊢ L = Line 𝒢 ⁡ G
colperpex.g ⊢ φ → G ∈ 𝒢 Tarski
perprag.1 ⊢ φ → A ∈ P
perprag.2 ⊢ φ → B ∈ P
perprag.3 ⊢ φ → C ∈ A L B
perprag.4 ⊢ φ → D ∈ P
perprag.5 ⊢ φ → A L B ⟂ 𝒢 ⁡ G C L D
Assertion perprag ⊢ φ → ⟨“ ACD ”⟩ ∈ ∟ 𝒢 ⁡ G

Proof

Step Hyp Ref Expression
1 colperpex.p ⊢ P = Base G
2 colperpex.d ⊢ - ˙ = dist ⁡ G
3 colperpex.i ⊢ I = Itv ⁡ G
4 colperpex.l ⊢ L = Line 𝒢 ⁡ G
5 colperpex.g ⊢ φ → G ∈ 𝒢 Tarski
6 perprag.1 ⊢ φ → A ∈ P
7 perprag.2 ⊢ φ → B ∈ P
8 perprag.3 ⊢ φ → C ∈ A L B
9 perprag.4 ⊢ φ → D ∈ P
10 perprag.5 ⊢ φ → A L B ⟂ 𝒢 ⁡ G C L D
11 eqidd ⊢ φ ∧ C = D → A = A
12 simpr ⊢ φ ∧ C = D → C = D
13 eqidd ⊢ φ ∧ C = D → D = D
14 11 12 13 s3eqd ⊢ φ ∧ C = D → ⟨“ ACD ”⟩ = ⟨“ ADD ”⟩
15 eqid ⊢ pInv 𝒢 ⁡ G = pInv 𝒢 ⁡ G
16 1 2 3 4 15 5 6 9 9 ragtrivb ⊢ φ → ⟨“ ADD ”⟩ ∈ ∟ 𝒢 ⁡ G
17 16 adantr ⊢ φ ∧ C = D → ⟨“ ADD ”⟩ ∈ ∟ 𝒢 ⁡ G
18 14 17 eqeltrd ⊢ φ ∧ C = D → ⟨“ ACD ”⟩ ∈ ∟ 𝒢 ⁡ G
19 5 adantr ⊢ φ ∧ C ≠ D → G ∈ 𝒢 Tarski
20 1 4 3 5 6 7 8 tglngne ⊢ φ → A ≠ B
21 1 3 4 5 6 7 20 tgelrnln ⊢ φ → A L B ∈ ran ⁡ L
22 21 adantr ⊢ φ ∧ C ≠ D → A L B ∈ ran ⁡ L
23 1 4 3 5 21 8 tglnpt ⊢ φ → C ∈ P
24 23 adantr ⊢ φ ∧ C ≠ D → C ∈ P
25 9 adantr ⊢ φ ∧ C ≠ D → D ∈ P
26 simpr ⊢ φ ∧ C ≠ D → C ≠ D
27 1 3 4 19 24 25 26 tgelrnln ⊢ φ ∧ C ≠ D → C L D ∈ ran ⁡ L
28 8 adantr ⊢ φ ∧ C ≠ D → C ∈ A L B
29 1 3 4 19 24 25 26 tglinerflx1 ⊢ φ ∧ C ≠ D → C ∈ C L D
30 28 29 elind ⊢ φ ∧ C ≠ D → C ∈ A L B ∩ C L D
31 1 3 4 5 6 7 20 tglinerflx1 ⊢ φ → A ∈ A L B
32 31 adantr ⊢ φ ∧ C ≠ D → A ∈ A L B
33 1 3 4 19 24 25 26 tglinerflx2 ⊢ φ ∧ C ≠ D → D ∈ C L D
34 10 adantr ⊢ φ ∧ C ≠ D → A L B ⟂ 𝒢 ⁡ G C L D
35 1 2 3 4 19 22 27 30 32 33 34 isperp2d ⊢ φ ∧ C ≠ D → ⟨“ ACD ”⟩ ∈ ∟ 𝒢 ⁡ G
36 18 35 pm2.61dane ⊢ φ → ⟨“ ACD ”⟩ ∈ ∟ 𝒢 ⁡ G