perpdrag

Description: Deduce a right angle from perpendicular lines. (Contributed by Thierry Arnoux, 12-Dec-2019)

	Ref	Expression
Hypotheses	colperpex.p	⊢ 𝑃 = ( Base ‘ 𝐺 )
	colperpex.d	⊢ − = ( dist ‘ 𝐺 )
	colperpex.i	⊢ 𝐼 = ( Itv ‘ 𝐺 )
	colperpex.l	⊢ 𝐿 = ( LineG ‘ 𝐺 )
	colperpex.g	⊢ ( 𝜑 → 𝐺 ∈ TarskiG )
	perpdrag.1	⊢ ( 𝜑 → 𝐴 ∈ 𝐷 )
	perpdrag.2	⊢ ( 𝜑 → 𝐵 ∈ 𝐷 )
	perpdrag.3	⊢ ( 𝜑 → 𝐶 ∈ 𝑃 )
	perpdrag.4	⊢ ( 𝜑 → 𝐷 ( ⟂G ‘ 𝐺 ) ( 𝐵 𝐿 𝐶 ) )
Assertion	perpdrag	⊢ ( 𝜑 → ⟨“ 𝐴 𝐵 𝐶 ”⟩ ∈ ( ∟G ‘ 𝐺 ) )

Proof

Step	Hyp	Ref	Expression
1		colperpex.p	⊢ 𝑃 = ( Base ‘ 𝐺 )
2		colperpex.d	⊢ − = ( dist ‘ 𝐺 )
3		colperpex.i	⊢ 𝐼 = ( Itv ‘ 𝐺 )
4		colperpex.l	⊢ 𝐿 = ( LineG ‘ 𝐺 )
5		colperpex.g	⊢ ( 𝜑 → 𝐺 ∈ TarskiG )
6		perpdrag.1	⊢ ( 𝜑 → 𝐴 ∈ 𝐷 )
7		perpdrag.2	⊢ ( 𝜑 → 𝐵 ∈ 𝐷 )
8		perpdrag.3	⊢ ( 𝜑 → 𝐶 ∈ 𝑃 )
9		perpdrag.4	⊢ ( 𝜑 → 𝐷 ( ⟂G ‘ 𝐺 ) ( 𝐵 𝐿 𝐶 ) )
10	5	ad2antrr	⊢ ( ( ( 𝜑 ∧ 𝑥 ∈ 𝐷 ) ∧ 𝐴 ≠ 𝑥 ) → 𝐺 ∈ TarskiG )
11	9	ad2antrr	⊢ ( ( ( 𝜑 ∧ 𝑥 ∈ 𝐷 ) ∧ 𝐴 ≠ 𝑥 ) → 𝐷 ( ⟂G ‘ 𝐺 ) ( 𝐵 𝐿 𝐶 ) )
12	4 10 11	perpln1	⊢ ( ( ( 𝜑 ∧ 𝑥 ∈ 𝐷 ) ∧ 𝐴 ≠ 𝑥 ) → 𝐷 ∈ ran 𝐿 )
13	6	ad2antrr	⊢ ( ( ( 𝜑 ∧ 𝑥 ∈ 𝐷 ) ∧ 𝐴 ≠ 𝑥 ) → 𝐴 ∈ 𝐷 )
14	1 4 3 10 12 13	tglnpt	⊢ ( ( ( 𝜑 ∧ 𝑥 ∈ 𝐷 ) ∧ 𝐴 ≠ 𝑥 ) → 𝐴 ∈ 𝑃 )
15		simplr	⊢ ( ( ( 𝜑 ∧ 𝑥 ∈ 𝐷 ) ∧ 𝐴 ≠ 𝑥 ) → 𝑥 ∈ 𝐷 )
16	1 4 3 10 12 15	tglnpt	⊢ ( ( ( 𝜑 ∧ 𝑥 ∈ 𝐷 ) ∧ 𝐴 ≠ 𝑥 ) → 𝑥 ∈ 𝑃 )
17	7	ad2antrr	⊢ ( ( ( 𝜑 ∧ 𝑥 ∈ 𝐷 ) ∧ 𝐴 ≠ 𝑥 ) → 𝐵 ∈ 𝐷 )
18		simpr	⊢ ( ( ( 𝜑 ∧ 𝑥 ∈ 𝐷 ) ∧ 𝐴 ≠ 𝑥 ) → 𝐴 ≠ 𝑥 )
19	1 3 4 10 14 16 18 18 12 13 15	tglinethru	⊢ ( ( ( 𝜑 ∧ 𝑥 ∈ 𝐷 ) ∧ 𝐴 ≠ 𝑥 ) → 𝐷 = ( 𝐴 𝐿 𝑥 ) )
20	17 19	eleqtrd	⊢ ( ( ( 𝜑 ∧ 𝑥 ∈ 𝐷 ) ∧ 𝐴 ≠ 𝑥 ) → 𝐵 ∈ ( 𝐴 𝐿 𝑥 ) )
21	8	ad2antrr	⊢ ( ( ( 𝜑 ∧ 𝑥 ∈ 𝐷 ) ∧ 𝐴 ≠ 𝑥 ) → 𝐶 ∈ 𝑃 )
22	19 11	eqbrtrrd	⊢ ( ( ( 𝜑 ∧ 𝑥 ∈ 𝐷 ) ∧ 𝐴 ≠ 𝑥 ) → ( 𝐴 𝐿 𝑥 ) ( ⟂G ‘ 𝐺 ) ( 𝐵 𝐿 𝐶 ) )
23	1 2 3 4 10 14 16 20 21 22	perprag	⊢ ( ( ( 𝜑 ∧ 𝑥 ∈ 𝐷 ) ∧ 𝐴 ≠ 𝑥 ) → ⟨“ 𝐴 𝐵 𝐶 ”⟩ ∈ ( ∟G ‘ 𝐺 ) )
24	4 5 9	perpln1	⊢ ( 𝜑 → 𝐷 ∈ ran 𝐿 )
25	1 3 4 5 24 6	tglnpt2	⊢ ( 𝜑 → ∃ 𝑥 ∈ 𝐷 𝐴 ≠ 𝑥 )
26	23 25	r19.29a	⊢ ( 𝜑 → ⟨“ 𝐴 𝐵 𝐶 ”⟩ ∈ ( ∟G ‘ 𝐺 ) )

Metamath Proof Explorer

Theorem perpdrag

Proof