Metamath Proof Explorer


Theorem cgrcgra

Description: Triangle congruence implies angle congruence. This is a portion of CPCTC, focusing on a specific angle. (Contributed by Arnoux, 2-Aug-2020)

Ref Expression
Hypotheses cgraid.p ⊢ P = Base G
cgraid.i ⊢ I = Itv ⁡ G
cgraid.g ⊢ φ → G ∈ 𝒢 Tarski
cgraid.k ⊢ K = hl 𝒢 ⁡ G
cgraid.a ⊢ φ → A ∈ P
cgraid.b ⊢ φ → B ∈ P
cgraid.c ⊢ φ → C ∈ P
cgracom.d ⊢ φ → D ∈ P
cgracom.e ⊢ φ → E ∈ P
cgracom.f ⊢ φ → F ∈ P
cgrcgra.1 ⊢ φ → A ≠ B
cgrcgra.2 ⊢ φ → B ≠ C
cgrcgra.3 ⊢ φ → ⟨“ ABC ”⟩ ∼ 𝒢 ⁡ G ⟨“ DEF ”⟩
Assertion cgrcgra ⊢ φ → ⟨“ ABC ”⟩ ∼ 𝒢 ∠ ⁡ G ⟨“ DEF ”⟩

Proof

Step Hyp Ref Expression
1 cgraid.p ⊢ P = Base G
2 cgraid.i ⊢ I = Itv ⁡ G
3 cgraid.g ⊢ φ → G ∈ 𝒢 Tarski
4 cgraid.k ⊢ K = hl 𝒢 ⁡ G
5 cgraid.a ⊢ φ → A ∈ P
6 cgraid.b ⊢ φ → B ∈ P
7 cgraid.c ⊢ φ → C ∈ P
8 cgracom.d ⊢ φ → D ∈ P
9 cgracom.e ⊢ φ → E ∈ P
10 cgracom.f ⊢ φ → F ∈ P
11 cgrcgra.1 ⊢ φ → A ≠ B
12 cgrcgra.2 ⊢ φ → B ≠ C
13 cgrcgra.3 ⊢ φ → ⟨“ ABC ”⟩ ∼ 𝒢 ⁡ G ⟨“ DEF ”⟩
14 eqid ⊢ dist ⁡ G = dist ⁡ G
15 eqid ⊢ ∼ 𝒢 ⁡ G = ∼ 𝒢 ⁡ G
16 1 14 2 15 3 5 6 7 8 9 10 13 cgr3simp1 ⊢ φ → A dist ⁡ G B = D dist ⁡ G E
17 1 14 2 3 5 6 8 9 16 11 tgcgrneq ⊢ φ → D ≠ E
18 1 2 4 8 5 9 3 17 hlid ⊢ φ → D K ⁡ E D
19 1 14 2 15 3 5 6 7 8 9 10 13 cgr3simp2 ⊢ φ → B dist ⁡ G C = E dist ⁡ G F
20 1 14 2 3 6 7 9 10 19 tgcgrcomlr ⊢ φ → C dist ⁡ G B = F dist ⁡ G E
21 12 necomd ⊢ φ → C ≠ B
22 1 14 2 3 7 6 10 9 20 21 tgcgrneq ⊢ φ → F ≠ E
23 1 2 4 10 5 9 3 22 hlid ⊢ φ → F K ⁡ E F
24 1 2 4 3 5 6 7 8 9 10 8 10 13 18 23 iscgrad ⊢ φ → ⟨“ ABC ”⟩ ∼ 𝒢 ∠ ⁡ G ⟨“ DEF ”⟩