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