Metamath Proof Explorer


Theorem ragflat2

Description: Deduce equality from two right angles. Theorem 8.6 of Schwabhauser p. 58. (Contributed by Thierry Arnoux, 3-Sep-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
ragflat2.d φ D P
ragflat2.1 φ ⟨“ ABC ”⟩ 𝒢 G
ragflat2.2 φ ⟨“ DBC ”⟩ 𝒢 G
ragflat2.3 φ C A I D
Assertion ragflat2 φ B = C

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 ragflat2.d φ D P
11 ragflat2.1 φ ⟨“ ABC ”⟩ 𝒢 G
12 ragflat2.2 φ ⟨“ DBC ”⟩ 𝒢 G
13 ragflat2.3 φ C A I D
14 eqid 𝒢 G = 𝒢 G
15 eqid S B = S B
16 1 2 3 4 5 6 8 15 9 mircl φ S B C P
17 1 2 3 4 5 6 7 8 9 israg φ ⟨“ ABC ”⟩ 𝒢 G A - ˙ C = A - ˙ S B C
18 11 17 mpbid φ A - ˙ C = A - ˙ S B C
19 1 2 3 4 5 6 10 8 9 israg φ ⟨“ DBC ”⟩ 𝒢 G D - ˙ C = D - ˙ S B C
20 12 19 mpbid φ D - ˙ C = D - ˙ S B C
21 1 4 3 6 7 10 9 14 16 7 2 13 18 20 tgidinside φ C = S B C
22 21 eqcomd φ S B C = C
23 1 2 3 4 5 6 8 15 9 mirinv φ S B C = C B = C
24 22 23 mpbid φ B = C