Metamath Proof Explorer


Theorem zerocgra

Description: Zero angles are congruent. Zero angles, that is, angles of degree zero, can be expressed by stating that points A and C are on the same ray starting at B , that is, A ( KB ) C . See also flatcgra . (Contributed by Thierry Arnoux, 23-Aug-2026)

Ref Expression
Hypotheses zerocgra.p P = Base G
zerocgra.a ˙ = 𝒢 G
zerocgra.k K = hl 𝒢 G
zerocgra.g φ G 𝒢 Tarski
zerocgra.1 φ A K B C
zerocgra.2 φ D K E F
zerocgra.b φ B P
zerocgra.e φ E P
Assertion zerocgra φ ⟨“ ABC ”⟩ ˙ ⟨“ DEF ”⟩

Proof

Step Hyp Ref Expression
1 zerocgra.p P = Base G
2 zerocgra.a ˙ = 𝒢 G
3 zerocgra.k K = hl 𝒢 G
4 zerocgra.g φ G 𝒢 Tarski
5 zerocgra.1 φ A K B C
6 zerocgra.2 φ D K E F
7 zerocgra.b φ B P
8 zerocgra.e φ E P
9 2 eqcomi 𝒢 G = ˙
10 9 a1i φ d P d K E D E dist G d = B dist G A f P f K E F E dist G f = B dist G C 𝒢 G = ˙
11 eqid Itv G = Itv G
12 4 ad6antr φ d P d K E D E dist G d = B dist G A f P f K E F E dist G f = B dist G C G 𝒢 Tarski
13 1 11 3 4 7 5 hlgrcl1 φ A P
14 13 ad6antr φ d P d K E D E dist G d = B dist G A f P f K E F E dist G f = B dist G C A P
15 7 ad6antr φ d P d K E D E dist G d = B dist G A f P f K E F E dist G f = B dist G C B P
16 1 11 3 4 7 5 hlgrcl2 φ C P
17 16 ad6antr φ d P d K E D E dist G d = B dist G A f P f K E F E dist G f = B dist G C C P
18 1 11 3 4 8 6 hlgrcl1 φ D P
19 18 ad6antr φ d P d K E D E dist G d = B dist G A f P f K E F E dist G f = B dist G C D P
20 8 ad6antr φ d P d K E D E dist G d = B dist G A f P f K E F E dist G f = B dist G C E P
21 1 11 3 4 8 6 hlgrcl2 φ F P
22 21 ad6antr φ d P d K E D E dist G d = B dist G A f P f K E F E dist G f = B dist G C F P
23 simp-6r φ d P d K E D E dist G d = B dist G A f P f K E F E dist G f = B dist G C d P
24 simpllr φ d P d K E D E dist G d = B dist G A f P f K E F E dist G f = B dist G C f P
25 eqid dist G = dist G
26 eqid 𝒢 G = 𝒢 G
27 simp-4r φ d P d K E D E dist G d = B dist G A f P f K E F E dist G f = B dist G C E dist G d = B dist G A
28 27 eqcomd φ d P d K E D E dist G d = B dist G A f P f K E F E dist G f = B dist G C B dist G A = E dist G d
29 1 25 11 12 15 14 20 23 28 tgcgrcomlr φ d P d K E D E dist G d = B dist G A f P f K E F E dist G f = B dist G C A dist G B = d dist G E
30 simpr φ d P d K E D E dist G d = B dist G A f P f K E F E dist G f = B dist G C E dist G f = B dist G C
31 30 eqcomd φ d P d K E D E dist G d = B dist G A f P f K E F E dist G f = B dist G C B dist G C = E dist G f
32 5 ad6antr φ d P d K E D E dist G d = B dist G A f P f K E F E dist G f = B dist G C A K B C
33 simp-5r φ d P d K E D E dist G d = B dist G A f P f K E F E dist G f = B dist G C d K E D
34 simplr φ d P d K E D E dist G d = B dist G A f P f K E F E dist G f = B dist G C f K E F
35 6 ad6antr φ d P d K E D E dist G d = B dist G A f P f K E F E dist G f = B dist G C D K E F
36 1 11 3 19 22 20 12 35 hlcomd φ d P d K E D E dist G d = B dist G A f P f K E F E dist G f = B dist G C F K E D
37 1 11 3 24 22 19 12 20 34 36 hltr φ d P d K E D E dist G d = B dist G A f P f K E F E dist G f = B dist G C f K E D
38 1 11 3 24 19 20 12 37 hlcomd φ d P d K E D E dist G d = B dist G A f P f K E F E dist G f = B dist G C D K E f
39 1 11 3 23 19 24 12 20 33 38 hltr φ d P d K E D E dist G d = B dist G A f P f K E F E dist G f = B dist G C d K E f
40 1 25 3 12 15 20 32 39 31 28 tghlsub φ d P d K E D E dist G d = B dist G A f P f K E F E dist G f = B dist G C C dist G A = f dist G d
41 1 25 26 12 14 15 17 23 20 24 29 31 40 trgcgr φ d P d K E D E dist G d = B dist G A f P f K E F E dist G f = B dist G C ⟨“ ABC ”⟩ 𝒢 G ⟨“ dEf ”⟩
42 1 11 3 12 14 15 17 19 20 22 23 24 41 33 34 iscgrad φ d P d K E D E dist G d = B dist G A f P f K E F E dist G f = B dist G C ⟨“ ABC ”⟩ 𝒢 G ⟨“ DEF ”⟩
43 10 42 breqdi φ d P d K E D E dist G d = B dist G A f P f K E F E dist G f = B dist G C ⟨“ ABC ”⟩ ˙ ⟨“ DEF ”⟩
44 43 anasss φ d P d K E D E dist G d = B dist G A f P f K E F E dist G f = B dist G C ⟨“ ABC ”⟩ ˙ ⟨“ DEF ”⟩
45 1 11 3 18 21 8 4 6 hlne2 φ F E
46 1 11 3 13 16 7 4 5 hlne2 φ C B
47 46 necomd φ B C
48 1 11 3 8 7 16 4 21 25 45 47 hlcgrex φ f P f K E F E dist G f = B dist G C
49 48 ad3antrrr φ d P d K E D E dist G d = B dist G A f P f K E F E dist G f = B dist G C
50 44 49 r19.29a φ d P d K E D E dist G d = B dist G A ⟨“ ABC ”⟩ ˙ ⟨“ DEF ”⟩
51 50 anasss φ d P d K E D E dist G d = B dist G A ⟨“ ABC ”⟩ ˙ ⟨“ DEF ”⟩
52 1 11 3 18 21 8 4 6 hlne1 φ D E
53 1 11 3 13 16 7 4 5 hlne1 φ A B
54 53 necomd φ B A
55 1 11 3 8 7 13 4 18 25 52 54 hlcgrex φ d P d K E D E dist G d = B dist G A
56 51 55 r19.29a φ ⟨“ ABC ”⟩ ˙ ⟨“ DEF ”⟩