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 = ( LineG ‘ 𝐺 )
	israg.s	⊢ 𝑆 = ( pInvG ‘ 𝐺 )
	israg.g	⊢ ( 𝜑 → 𝐺 ∈ TarskiG )
	israg.a	⊢ ( 𝜑 → 𝐴 ∈ 𝑃 )
	israg.b	⊢ ( 𝜑 → 𝐵 ∈ 𝑃 )
	israg.c	⊢ ( 𝜑 → 𝐶 ∈ 𝑃 )
	ragflat2.d	⊢ ( 𝜑 → 𝐷 ∈ 𝑃 )
	ragflat2.1	⊢ ( 𝜑 → ⟨“ 𝐴 𝐵 𝐶 ”⟩ ∈ ( ∟G ‘ 𝐺 ) )
	ragflat2.2	⊢ ( 𝜑 → ⟨“ 𝐷 𝐵 𝐶 ”⟩ ∈ ( ∟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 = ( LineG ‘ 𝐺 )
5		israg.s	⊢ 𝑆 = ( pInvG ‘ 𝐺 )
6		israg.g	⊢ ( 𝜑 → 𝐺 ∈ TarskiG )
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 → ⟨“ 𝐴 𝐵 𝐶 ”⟩ ∈ ( ∟G ‘ 𝐺 ) )
12		ragflat2.2	⊢ ( 𝜑 → ⟨“ 𝐷 𝐵 𝐶 ”⟩ ∈ ( ∟G ‘ 𝐺 ) )
13		ragflat2.3	⊢ ( 𝜑 → 𝐶 ∈ ( 𝐴 𝐼 𝐷 ) )
14		eqid	⊢ ( cgrG ‘ 𝐺 ) = ( cgrG ‘ 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 → ( ⟨“ 𝐴 𝐵 𝐶 ”⟩ ∈ ( ∟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 → ( ⟨“ 𝐷 𝐵 𝐶 ”⟩ ∈ ( ∟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 )