Metamath Proof Explorer


Theorem inagflat

Description: Any point lies in a flat angle. (Contributed by Thierry Arnoux, 13-Feb-2023)

Ref Expression
Hypotheses isinag.p P = Base G
isinag.i I = Itv G
isinag.k K = hl 𝒢 G
isinag.x φ X P
isinag.a φ A P
isinag.b φ B P
isinag.c φ C P
inagflat.g φ G 𝒢 Tarski
inagflat.x φ X P
inagflat.1 φ A B
inagflat.2 φ C B
inagflat.3 φ X B
inagflat.4 φ B A I C
Assertion inagflat φ X 𝒢 G ⟨“ ABC ”⟩

Proof

Step Hyp Ref Expression
1 isinag.p P = Base G
2 isinag.i I = Itv G
3 isinag.k K = hl 𝒢 G
4 isinag.x φ X P
5 isinag.a φ A P
6 isinag.b φ B P
7 isinag.c φ C P
8 inagflat.g φ G 𝒢 Tarski
9 inagflat.x φ X P
10 inagflat.1 φ A B
11 inagflat.2 φ C B
12 inagflat.3 φ X B
13 inagflat.4 φ B A I C
14 eqidd φ B = B
15 14 orcd φ B = B B K B X
16 1 2 3 4 5 6 7 8 6 10 11 12 13 15 isinagd φ X 𝒢 G ⟨“ ABC ”⟩