Metamath Proof Explorer


Theorem tgextaoa

Description: The two exterior angles at C are VERTICAL, hence congruent. Ray C D is opposite ray C B and ray C E is opposite ray C A , so <" B C E "> and <" A C D "> have both legs flipped. One sacgr (Schwabhaeuser 11.13, GeoCoq l11__13 ) does it in a single shot, because sacgr flips the first slot of each side independently: the premise it wants is <" D C E "> ( cgrAG ) <" E C D "> , a bare angle reversal by cgraswap , needing no non-collinearity at all. oacgr (11.14) would prove the same thing from two supplements and is not used. (Contributed by JAG, 12-Aug-2026)

Ref Expression
Hypotheses tgextaoa.p ⊢ P = Base G
tgextaoa.m ⊢ - ˙ = dist ⁡ G
tgextaoa.i ⊢ I = Itv ⁡ G
tgextaoa.k ⊢ K = hl 𝒢 ⁡ G
tgextaoa.g ⊢ φ → G ∈ 𝒢 Tarski
tgextaoa.a ⊢ φ → A ∈ P
tgextaoa.b ⊢ φ → B ∈ P
tgextaoa.c ⊢ φ → C ∈ P
tgextaoa.d ⊢ φ → D ∈ P
tgextaoa.e ⊢ φ → E ∈ P
tgextaoa.ext ⊢ φ → C ∈ B I D
tgextaoa.ext2 ⊢ φ → C ∈ A I E
tgextaoa.cb ⊢ φ → C ≠ B
tgextaoa.ca ⊢ φ → C ≠ A
tgextaoa.dc ⊢ φ → D ≠ C
tgextaoa.ce ⊢ φ → C ≠ E
Assertion tgextaoa ⊢ φ → ⟨“ BCE ”⟩ ∼ 𝒢 ∠ ⁡ G ⟨“ ACD ”⟩

Proof

Step Hyp Ref Expression
1 tgextaoa.p ⊢ P = Base G
2 tgextaoa.m ⊢ - ˙ = dist ⁡ G
3 tgextaoa.i ⊢ I = Itv ⁡ G
4 tgextaoa.k ⊢ K = hl 𝒢 ⁡ G
5 tgextaoa.g ⊢ φ → G ∈ 𝒢 Tarski
6 tgextaoa.a ⊢ φ → A ∈ P
7 tgextaoa.b ⊢ φ → B ∈ P
8 tgextaoa.c ⊢ φ → C ∈ P
9 tgextaoa.d ⊢ φ → D ∈ P
10 tgextaoa.e ⊢ φ → E ∈ P
11 tgextaoa.ext ⊢ φ → C ∈ B I D
12 tgextaoa.ext2 ⊢ φ → C ∈ A I E
13 tgextaoa.cb ⊢ φ → C ≠ B
14 tgextaoa.ca ⊢ φ → C ≠ A
15 tgextaoa.dc ⊢ φ → D ≠ C
16 tgextaoa.ce ⊢ φ → C ≠ E
17 1 3 5 4 9 8 10 15 16 cgraswap ⊢ φ → ⟨“ DCE ”⟩ ∼ 𝒢 ∠ ⁡ G ⟨“ ECD ”⟩
18 1 2 3 5 7 8 9 11 tgbtwncom ⊢ φ → C ∈ D I B
19 1 2 3 5 6 8 10 12 tgbtwncom ⊢ φ → C ∈ E I A
20 1 3 2 5 9 8 10 10 8 9 7 6 17 18 19 13 14 sacgr ⊢ φ → ⟨“ BCE ”⟩ ∼ 𝒢 ∠ ⁡ G ⟨“ ACD ”⟩