Metamath Proof Explorer


Theorem chordthmALT

Description: The intersecting chords theorem. If points A, B, C, and D lie on a circle (with center Q, say), and the point P is on the interior of the segments AB and CD, then the two products of lengths PA x. PB and PC x. PD are equal. The Euclidean plane is identified with the complex plane, and the fact that P is on AB and on CD is expressed by the hypothesis that the angles APB and CPD are equal to _pi . The result is proven by using chordthmlem5 twice to show that PA x. PB and PC x. PD both equal BQ2 - PQ2. This is similar to the proof of the theorem given in Euclid'sElements, where it is Proposition III.35. Proven by David Moews on 28-Feb-2017 as chordthm . https://us.metamath.org/other/completeusersproof/chordthmaltvd.html is a Virtual Deduction User's Proof transcription of chordthm . That VD User's Proof was input into completeusersproof, automatically generating this chordthmALT Metamath proof. (Contributed by Alan Sare, 19-Sep-2017) (Proof modification is discouraged.) (New usage is discouraged.)

Ref Expression
Hypotheses chordthmALT.angdef F=x0,y0logyx
chordthmALT.A φA
chordthmALT.B φB
chordthmALT.C φC
chordthmALT.D φD
chordthmALT.P φP
chordthmALT.AneP φAP
chordthmALT.BneP φBP
chordthmALT.CneP φCP
chordthmALT.DneP φDP
chordthmALT.APB φAPFBP=π
chordthmALT.CPD φCPFDP=π
chordthmALT.Q φQ
chordthmALT.ABcirc φAQ=BQ
chordthmALT.ACcirc φAQ=CQ
chordthmALT.ADcirc φAQ=DQ
Assertion chordthmALT φPAPB=PCPD

Proof

Step Hyp Ref Expression
1 chordthmALT.angdef F=x0,y0logyx
2 chordthmALT.A φA
3 chordthmALT.B φB
4 chordthmALT.C φC
5 chordthmALT.D φD
6 chordthmALT.P φP
7 chordthmALT.AneP φAP
8 chordthmALT.BneP φBP
9 chordthmALT.CneP φCP
10 chordthmALT.DneP φDP
11 chordthmALT.APB φAPFBP=π
12 chordthmALT.CPD φCPFDP=π
13 chordthmALT.Q φQ
14 chordthmALT.ABcirc φAQ=BQ
15 chordthmALT.ACcirc φAQ=CQ
16 chordthmALT.ADcirc φAQ=DQ
17 10 necomd φPD
18 1 4 6 5 9 17 angpieqvd φCPFDP=πv01P=vC+1vD
19 12 18 mpbid φv01P=vC+1vD
20 df-rex v01P=vC+1vDvv01P=vC+1vD
21 20 biimpi v01P=vC+1vDvv01P=vC+1vD
22 19 21 syl φvv01P=vC+1vD
23 8 necomd φPB
24 1 2 6 3 7 23 angpieqvd φAPFBP=πw01P=wA+1wB
25 11 24 mpbid φw01P=wA+1wB
26 df-rex w01P=wA+1wBww01P=wA+1wB
27 26 biimpi w01P=wA+1wBww01P=wA+1wB
28 25 27 syl φww01P=wA+1wB
29 28 adantr φv01P=vC+1vDww01P=wA+1wB
30 14 16 eqtr3d φBQ=DQ
31 30 oveq1d φBQ2=DQ2
32 31 oveq1d φBQ2PQ2=DQ2PQ2
33 32 3ad2ant1 φv01P=vC+1vDw01P=wA+1wBBQ2PQ2=DQ2PQ2
34 2 3ad2ant1 φw01P=wA+1wBA
35 3 3ad2ant1 φw01P=wA+1wBB
36 13 3ad2ant1 φw01P=wA+1wBQ
37 ioossicc 0101
38 id w01w01
39 37 38 sselid w01w01
40 39 3ad2ant2 φw01P=wA+1wBw01
41 id P=wA+1wBP=wA+1wB
42 41 3ad2ant3 φw01P=wA+1wBP=wA+1wB
43 14 3ad2ant1 φw01P=wA+1wBAQ=BQ
44 34 35 36 40 42 43 chordthmlem5 φw01P=wA+1wBPAPB=BQ2PQ2
45 44 3expb φw01P=wA+1wBPAPB=BQ2PQ2
46 45 3adant2 φv01P=vC+1vDw01P=wA+1wBPAPB=BQ2PQ2
47 4 3ad2ant1 φv01P=vC+1vDC
48 5 3ad2ant1 φv01P=vC+1vDD
49 13 3ad2ant1 φv01P=vC+1vDQ
50 id v01v01
51 37 50 sselid v01v01
52 51 3ad2ant2 φv01P=vC+1vDv01
53 id P=vC+1vDP=vC+1vD
54 53 3ad2ant3 φv01P=vC+1vDP=vC+1vD
55 15 16 eqtr3d φCQ=DQ
56 55 3ad2ant1 φv01P=vC+1vDCQ=DQ
57 47 48 49 52 54 56 chordthmlem5 φv01P=vC+1vDPCPD=DQ2PQ2
58 57 3expb φv01P=vC+1vDPCPD=DQ2PQ2
59 58 3adant3 φv01P=vC+1vDw01P=wA+1wBPCPD=DQ2PQ2
60 33 46 59 3eqtr4d φv01P=vC+1vDw01P=wA+1wBPAPB=PCPD
61 60 3expia φv01P=vC+1vDw01P=wA+1wBPAPB=PCPD
62 61 exlimdv φv01P=vC+1vDww01P=wA+1wBPAPB=PCPD
63 29 62 mpd φv01P=vC+1vDPAPB=PCPD
64 63 ex φv01P=vC+1vDPAPB=PCPD
65 64 exlimdv φvv01P=vC+1vDPAPB=PCPD
66 22 65 mpd φPAPB=PCPD