Metamath Proof Explorer


Theorem lmimid

Description: If we have a right angle, then the mirror point is the point inversion. (Contributed by Thierry Arnoux, 15-Dec-2019)

Ref Expression
Hypotheses ismid.p P=BaseG
ismid.d -˙=distG
ismid.i I=ItvG
ismid.g φG𝒢Tarski
ismid.1 φGDim𝒢2
lmif.m M=lInv𝒢GD
lmif.l L=Line𝒢G
lmif.d φDranL
lmicl.1 φAP
lmimid.s S=pInv𝒢GB
lmimid.r φ⟨“ABC”⟩𝒢G
lmimid.a φAD
lmimid.b φBD
lmimid.c φCP
lmimid.d φAB
Assertion lmimid φMC=SC

Proof

Step Hyp Ref Expression
1 ismid.p P=BaseG
2 ismid.d -˙=distG
3 ismid.i I=ItvG
4 ismid.g φG𝒢Tarski
5 ismid.1 φGDim𝒢2
6 lmif.m M=lInv𝒢GD
7 lmif.l L=Line𝒢G
8 lmif.d φDranL
9 lmicl.1 φAP
10 lmimid.s S=pInv𝒢GB
11 lmimid.r φ⟨“ABC”⟩𝒢G
12 lmimid.a φAD
13 lmimid.b φBD
14 lmimid.c φCP
15 lmimid.d φAB
16 10 a1i φS=pInv𝒢GB
17 16 fveq1d φSC=pInv𝒢GBC
18 eqid pInv𝒢G=pInv𝒢G
19 1 7 3 4 8 13 tglnpt φBP
20 1 2 3 7 18 4 19 10 14 mircl φSCP
21 1 2 3 4 5 14 20 18 19 ismidb φSC=pInv𝒢GBCCmid𝒢GSC=B
22 17 21 mpbid φCmid𝒢GSC=B
23 22 13 eqeltrd φCmid𝒢GSCD
24 df-ne CSC¬C=SC
25 4 adantr φCSCG𝒢Tarski
26 8 adantr φCSCDranL
27 14 adantr φCSCCP
28 20 adantr φCSCSCP
29 simpr φCSCCSC
30 1 3 7 25 27 28 29 tgelrnln φCSCCLSCranL
31 13 adantr φCSCBD
32 19 adantr φCSCBP
33 1 2 3 4 5 14 20 midbtwn φCmid𝒢GSCCISC
34 22 33 eqeltrrd φBCISC
35 34 adantr φCSCBCISC
36 1 3 7 25 27 28 32 29 35 btwnlng1 φCSCBCLSC
37 31 36 elind φCSCBDCLSC
38 12 adantr φCSCAD
39 1 3 7 25 27 28 29 tglinerflx1 φCSCCCLSC
40 15 adantr φCSCAB
41 1 2 3 7 18 4 19 10 14 mirinv φSC=CB=C
42 eqcom B=CC=B
43 41 42 bitrdi φSC=CC=B
44 43 biimpar φC=BSC=C
45 44 eqcomd φC=BC=SC
46 45 ex φC=BC=SC
47 46 necon3d φCSCCB
48 47 imp φCSCCB
49 11 adantr φCSC⟨“ABC”⟩𝒢G
50 1 2 3 7 25 26 30 37 38 39 40 48 49 ragperp φCSCD𝒢GCLSC
51 50 ex φCSCD𝒢GCLSC
52 24 51 biimtrrid φ¬C=SCD𝒢GCLSC
53 52 orrd φC=SCD𝒢GCLSC
54 53 orcomd φD𝒢GCLSCC=SC
55 1 2 3 4 5 6 7 8 14 20 islmib φSC=MCCmid𝒢GSCDD𝒢GCLSCC=SC
56 23 54 55 mpbir2and φSC=MC
57 56 eqcomd φMC=SC