Metamath Proof Explorer


Theorem perpdragALT

Description: Deduce a right angle from perpendicular lines. (Contributed by Thierry Arnoux, 12-Dec-2019) (New usage is discouraged.) (Proof modification is discouraged.)

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
perpdrag.1 ⊢ φ → A ∈ D
perpdrag.2 ⊢ φ → B ∈ D
perpdrag.3 ⊢ φ → C ∈ P
perpdrag.4 ⊢ φ → D ⟂ 𝒢 ⁡ G B L C
Assertion perpdragALT ⊢ φ → ⟨“ ABC ”⟩ ∈ ∟ 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 perpdrag.1 ⊢ φ → A ∈ D
7 perpdrag.2 ⊢ φ → B ∈ D
8 perpdrag.3 ⊢ φ → C ∈ P
9 perpdrag.4 ⊢ φ → D ⟂ 𝒢 ⁡ G B L C
10 eqidd ⊢ φ ∧ A = B → A = A
11 simpr ⊢ φ ∧ A = B → A = B
12 eqidd ⊢ φ ∧ A = B → C = C
13 10 11 12 s3eqd ⊢ φ ∧ A = B → ⟨“ AAC ”⟩ = ⟨“ ABC ”⟩
14 eqid ⊢ pInv 𝒢 ⁡ G = pInv 𝒢 ⁡ G
15 4 5 9 perpln1 ⊢ φ → D ∈ ran ⁡ L
16 1 4 3 5 15 6 tglnpt ⊢ φ → A ∈ P
17 1 2 3 4 14 5 8 16 8 ragtrivb ⊢ φ → ⟨“ CAA ”⟩ ∈ ∟ 𝒢 ⁡ G
18 1 2 3 4 14 5 8 16 16 17 ragcom ⊢ φ → ⟨“ AAC ”⟩ ∈ ∟ 𝒢 ⁡ G
19 18 adantr ⊢ φ ∧ A = B → ⟨“ AAC ”⟩ ∈ ∟ 𝒢 ⁡ G
20 13 19 eqeltrrd ⊢ φ ∧ A = B → ⟨“ ABC ”⟩ ∈ ∟ 𝒢 ⁡ G
21 5 adantr ⊢ φ ∧ A ≠ B → G ∈ 𝒢 Tarski
22 16 adantr ⊢ φ ∧ A ≠ B → A ∈ P
23 1 4 3 5 15 7 tglnpt ⊢ φ → B ∈ P
24 23 adantr ⊢ φ ∧ A ≠ B → B ∈ P
25 7 adantr ⊢ φ ∧ A ≠ B → B ∈ D
26 simpr ⊢ φ ∧ A ≠ B → A ≠ B
27 15 adantr ⊢ φ ∧ A ≠ B → D ∈ ran ⁡ L
28 6 adantr ⊢ φ ∧ A ≠ B → A ∈ D
29 1 3 4 21 22 24 26 26 27 28 25 tglinethru ⊢ φ ∧ A ≠ B → D = A L B
30 25 29 eleqtrd ⊢ φ ∧ A ≠ B → B ∈ A L B
31 8 adantr ⊢ φ ∧ A ≠ B → C ∈ P
32 9 adantr ⊢ φ ∧ A ≠ B → D ⟂ 𝒢 ⁡ G B L C
33 29 32 eqbrtrrd ⊢ φ ∧ A ≠ B → A L B ⟂ 𝒢 ⁡ G B L C
34 1 2 3 4 21 22 24 30 31 33 perprag ⊢ φ ∧ A ≠ B → ⟨“ ABC ”⟩ ∈ ∟ 𝒢 ⁡ G
35 20 34 pm2.61dane ⊢ φ → ⟨“ ABC ”⟩ ∈ ∟ 𝒢 ⁡ G