Metamath Proof Explorer


Theorem mirrag

Description: Right angle is conserved by point inversion. (Contributed by Thierry Arnoux, 3-Nov-2019)

Ref Expression
Hypotheses israg.p ⊢ P = Base G
israg.d ⊢ - ˙ = dist ⁡ G
israg.i ⊢ I = Itv ⁡ G
israg.l ⊢ L = Line 𝒢 ⁡ G
israg.s ⊢ S = pInv 𝒢 ⁡ G
israg.g ⊢ φ → G ∈ 𝒢 Tarski
israg.a ⊢ φ → A ∈ P
israg.b ⊢ φ → B ∈ P
israg.c ⊢ φ → C ∈ P
ragmir.1 ⊢ φ → ⟨“ ABC ”⟩ ∈ ∟ 𝒢 ⁡ G
mirrag.m ⊢ M = S ⁡ D
mirrag.d ⊢ φ → D ∈ P
Assertion mirrag ⊢ φ → ⟨“ M ⁡ A M ⁡ B M ⁡ C ”⟩ ∈ ∟ 𝒢 ⁡ G

Proof

Step Hyp Ref Expression
1 israg.p ⊢ P = Base G
2 israg.d ⊢ - ˙ = dist ⁡ G
3 israg.i ⊢ I = Itv ⁡ G
4 israg.l ⊢ L = Line 𝒢 ⁡ G
5 israg.s ⊢ S = pInv 𝒢 ⁡ G
6 israg.g ⊢ φ → G ∈ 𝒢 Tarski
7 israg.a ⊢ φ → A ∈ P
8 israg.b ⊢ φ → B ∈ P
9 israg.c ⊢ φ → C ∈ P
10 ragmir.1 ⊢ φ → ⟨“ ABC ”⟩ ∈ ∟ 𝒢 ⁡ G
11 mirrag.m ⊢ M = S ⁡ D
12 mirrag.d ⊢ φ → D ∈ P
13 eqid ⊢ S ⁡ B = S ⁡ B
14 1 2 3 4 5 6 8 13 9 mircl ⊢ φ → S ⁡ B ⁡ C ∈ P
15 1 2 3 4 5 6 7 8 9 israg ⊢ φ → ⟨“ ABC ”⟩ ∈ ∟ 𝒢 ⁡ G ↔ A - ˙ C = A - ˙ S ⁡ B ⁡ C
16 10 15 mpbid ⊢ φ → A - ˙ C = A - ˙ S ⁡ B ⁡ C
17 1 2 3 4 5 6 12 11 7 9 7 14 16 mircgrs ⊢ φ → M ⁡ A - ˙ M ⁡ C = M ⁡ A - ˙ M ⁡ S ⁡ B ⁡ C
18 1 2 3 4 5 6 12 11 9 8 mirmir2 ⊢ φ → M ⁡ S ⁡ B ⁡ C = S ⁡ M ⁡ B ⁡ M ⁡ C
19 18 oveq2d ⊢ φ → M ⁡ A - ˙ M ⁡ S ⁡ B ⁡ C = M ⁡ A - ˙ S ⁡ M ⁡ B ⁡ M ⁡ C
20 17 19 eqtrd ⊢ φ → M ⁡ A - ˙ M ⁡ C = M ⁡ A - ˙ S ⁡ M ⁡ B ⁡ M ⁡ C
21 1 2 3 4 5 6 12 11 7 mircl ⊢ φ → M ⁡ A ∈ P
22 1 2 3 4 5 6 12 11 8 mircl ⊢ φ → M ⁡ B ∈ P
23 1 2 3 4 5 6 12 11 9 mircl ⊢ φ → M ⁡ C ∈ P
24 1 2 3 4 5 6 21 22 23 israg ⊢ φ → ⟨“ M ⁡ A M ⁡ B M ⁡ C ”⟩ ∈ ∟ 𝒢 ⁡ G ↔ M ⁡ A - ˙ M ⁡ C = M ⁡ A - ˙ S ⁡ M ⁡ B ⁡ M ⁡ C
25 20 24 mpbird ⊢ φ → ⟨“ M ⁡ A M ⁡ B M ⁡ C ”⟩ ∈ ∟ 𝒢 ⁡ G