Metamath Proof Explorer


Theorem perprag

Description: Deduce a right angle from perpendicular lines. (Contributed by Thierry Arnoux, 10-Nov-2019)

Ref Expression
Hypotheses colperpex.p P = Base G
colperpex.d - ˙ = dist G
colperpex.i I = Itv G
colperpex.l L = Line 𝒢 G
colperpex.g φ G 𝒢 Tarski
perprag.1 φ A P
perprag.2 φ B P
perprag.3 φ C A L B
perprag.4 φ D P
perprag.5 φ A L B 𝒢 G C L D
Assertion perprag φ ⟨“ ACD ”⟩ 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 = Line 𝒢 G
5 colperpex.g φ G 𝒢 Tarski
6 perprag.1 φ A P
7 perprag.2 φ B P
8 perprag.3 φ C A L B
9 perprag.4 φ D P
10 perprag.5 φ A L B 𝒢 G C L D
11 eqidd φ C = D A = A
12 simpr φ C = D C = D
13 eqidd φ C = D D = D
14 11 12 13 s3eqd φ C = D ⟨“ ACD ”⟩ = ⟨“ ADD ”⟩
15 eqid pInv 𝒢 G = pInv 𝒢 G
16 1 2 3 4 15 5 6 9 9 ragtrivb φ ⟨“ ADD ”⟩ 𝒢 G
17 16 adantr φ C = D ⟨“ ADD ”⟩ 𝒢 G
18 14 17 eqeltrd φ C = D ⟨“ ACD ”⟩ 𝒢 G
19 5 adantr φ C D G 𝒢 Tarski
20 1 4 3 5 6 7 8 tglngne φ A B
21 1 3 4 5 6 7 20 tgelrnln φ A L B ran L
22 21 adantr φ C D A L B ran L
23 1 4 3 5 21 8 tglnpt φ C P
24 23 adantr φ C D C P
25 9 adantr φ C D D P
26 simpr φ C D C D
27 1 3 4 19 24 25 26 tgelrnln φ C D C L D ran L
28 8 adantr φ C D C A L B
29 1 3 4 19 24 25 26 tglinerflx1 φ C D C C L D
30 28 29 elind φ C D C A L B C L D
31 1 3 4 5 6 7 20 tglinerflx1 φ A A L B
32 31 adantr φ C D A A L B
33 1 3 4 19 24 25 26 tglinerflx2 φ C D D C L D
34 10 adantr φ C D A L B 𝒢 G C L D
35 1 2 3 4 19 22 27 30 32 33 34 isperp2d φ C D ⟨“ ACD ”⟩ 𝒢 G
36 18 35 pm2.61dane φ ⟨“ ACD ”⟩ 𝒢 G