Metamath Proof Explorer


Theorem perpdrag

Description: Deduce a right angle from perpendicular lines. (Contributed by Thierry Arnoux, 12-Dec-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
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 perpdrag ⊢ 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 5 ad2antrr ⊢ φ ∧ x ∈ D ∧ A ≠ x → G ∈ 𝒢 Tarski
11 9 ad2antrr ⊢ φ ∧ x ∈ D ∧ A ≠ x → D ⟂ 𝒢 ⁡ G B L C
12 4 10 11 perpln1 ⊢ φ ∧ x ∈ D ∧ A ≠ x → D ∈ ran ⁡ L
13 6 ad2antrr ⊢ φ ∧ x ∈ D ∧ A ≠ x → A ∈ D
14 1 4 3 10 12 13 tglnpt ⊢ φ ∧ x ∈ D ∧ A ≠ x → A ∈ P
15 simplr ⊢ φ ∧ x ∈ D ∧ A ≠ x → x ∈ D
16 1 4 3 10 12 15 tglnpt ⊢ φ ∧ x ∈ D ∧ A ≠ x → x ∈ P
17 7 ad2antrr ⊢ φ ∧ x ∈ D ∧ A ≠ x → B ∈ D
18 simpr ⊢ φ ∧ x ∈ D ∧ A ≠ x → A ≠ x
19 1 3 4 10 14 16 18 18 12 13 15 tglinethru ⊢ φ ∧ x ∈ D ∧ A ≠ x → D = A L x
20 17 19 eleqtrd ⊢ φ ∧ x ∈ D ∧ A ≠ x → B ∈ A L x
21 8 ad2antrr ⊢ φ ∧ x ∈ D ∧ A ≠ x → C ∈ P
22 19 11 eqbrtrrd ⊢ φ ∧ x ∈ D ∧ A ≠ x → A L x ⟂ 𝒢 ⁡ G B L C
23 1 2 3 4 10 14 16 20 21 22 perprag ⊢ φ ∧ x ∈ D ∧ A ≠ x → ⟨“ ABC ”⟩ ∈ ∟ 𝒢 ⁡ G
24 4 5 9 perpln1 ⊢ φ → D ∈ ran ⁡ L
25 1 3 4 5 24 6 tglnpt2 ⊢ φ → ∃ x ∈ D A ≠ x
26 23 25 r19.29a ⊢ φ → ⟨“ ABC ”⟩ ∈ ∟ 𝒢 ⁡ G