Metamath Proof Explorer


Theorem chordthmlem4

Description: If P is on the segment AB and M is the midpoint of AB, then PA x. PB = BM2 - PM2. If all lengths are reexpressed as fractions of AB, this reduces to the identity X x. ( 1 - X ) = ( 1 / 2 )2 - ( ( 1 / 2 ) - X )2. (Contributed by David Moews, 28-Feb-2017)

Ref Expression
Hypotheses chordthmlem4.A φ A
chordthmlem4.B φ B
chordthmlem4.X φ X 0 1
chordthmlem4.M φ M = A + B 2
chordthmlem4.P φ P = X A + 1 X B
Assertion chordthmlem4 φ P A P B = B M 2 P M 2

Proof

Step Hyp Ref Expression
1 chordthmlem4.A φ A
2 chordthmlem4.B φ B
3 chordthmlem4.X φ X 0 1
4 chordthmlem4.M φ M = A + B 2
5 chordthmlem4.P φ P = X A + 1 X B
6 1red φ 1
7 unitssre 0 1
8 7 3 sselid φ X
9 6 8 resubcld φ 1 X
10 9 recnd φ 1 X
11 10 abscld φ 1 X
12 11 recnd φ 1 X
13 2 1 subcld φ B A
14 13 abscld φ B A
15 14 recnd φ B A
16 8 recnd φ X
17 16 abscld φ X
18 17 recnd φ X
19 12 15 18 15 mul4d φ 1 X B A X B A = 1 X X B A B A
20 16 1 mulcld φ X A
21 10 2 mulcld φ 1 X B
22 20 21 addcld φ X A + 1 X B
23 5 22 eqeltrd φ P
24 1 23 2 16 affineequiv2 φ P = X A + 1 X B P A = 1 X B A
25 5 24 mpbid φ P A = 1 X B A
26 25 fveq2d φ P A = 1 X B A
27 10 13 absmuld φ 1 X B A = 1 X B A
28 26 27 eqtrd φ P A = 1 X B A
29 23 2 abssubd φ P B = B P
30 1 23 2 16 affineequiv φ P = X A + 1 X B B P = X B A
31 5 30 mpbid φ B P = X B A
32 31 fveq2d φ B P = X B A
33 16 13 absmuld φ X B A = X B A
34 29 32 33 3eqtrd φ P B = X B A
35 28 34 oveq12d φ P A P B = 1 X B A X B A
36 15 sqvald φ B A 2 = B A B A
37 36 oveq2d φ 1 X X B A 2 = 1 X X B A B A
38 19 35 37 3eqtr4d φ P A P B = 1 X X B A 2
39 6 recnd φ 1
40 39 halfcld φ 1 2
41 40 sqcld φ 1 2 2
42 6 rehalfcld φ 1 2
43 42 8 resubcld φ 1 2 X
44 43 recnd φ 1 2 X
45 44 abscld φ 1 2 X
46 45 recnd φ 1 2 X
47 46 sqcld φ 1 2 X 2
48 15 sqcld φ B A 2
49 41 47 48 subdird φ 1 2 2 1 2 X 2 B A 2 = 1 2 2 B A 2 1 2 X 2 B A 2
50 subsq 1 2 1 2 X 1 2 2 1 2 X 2 = 1 2 + 1 2 - X 1 2 1 2 X
51 40 44 50 syl2anc φ 1 2 2 1 2 X 2 = 1 2 + 1 2 - X 1 2 1 2 X
52 40 40 16 addsubassd φ 1 2 + 1 2 - X = 1 2 + 1 2 - X
53 39 2halvesd φ 1 2 + 1 2 = 1
54 53 oveq1d φ 1 2 + 1 2 - X = 1 X
55 52 54 eqtr3d φ 1 2 + 1 2 - X = 1 X
56 40 16 nncand φ 1 2 1 2 X = X
57 55 56 oveq12d φ 1 2 + 1 2 - X 1 2 1 2 X = 1 X X
58 51 57 eqtr2d φ 1 X X = 1 2 2 1 2 X 2
59 elicc01 X 0 1 X 0 X X 1
60 3 59 sylib φ X 0 X X 1
61 60 simp3d φ X 1
62 8 6 61 abssubge0d φ 1 X = 1 X
63 60 simp2d φ 0 X
64 8 63 absidd φ X = X
65 62 64 oveq12d φ 1 X X = 1 X X
66 absresq 1 2 X 1 2 X 2 = 1 2 X 2
67 43 66 syl φ 1 2 X 2 = 1 2 X 2
68 67 oveq2d φ 1 2 2 1 2 X 2 = 1 2 2 1 2 X 2
69 58 65 68 3eqtr4d φ 1 X X = 1 2 2 1 2 X 2
70 69 oveq1d φ 1 X X B A 2 = 1 2 2 1 2 X 2 B A 2
71 2cnd φ 2
72 2ne0 2 0
73 72 a1i φ 2 0
74 2 71 73 divcan4d φ B 2 2 = B
75 2 times2d φ B 2 = B + B
76 75 oveq1d φ B 2 2 = B + B 2
77 74 76 eqtr3d φ B = B + B 2
78 77 4 oveq12d φ B M = B + B 2 A + B 2
79 2 2 addcld φ B + B
80 1 2 addcld φ A + B
81 79 80 71 73 divsubdird φ B + B - A + B 2 = B + B 2 A + B 2
82 2 1 2 pnpcan2d φ B + B - A + B = B A
83 82 oveq1d φ B + B - A + B 2 = B A 2
84 78 81 83 3eqtr2d φ B M = B A 2
85 13 71 73 divrec2d φ B A 2 = 1 2 B A
86 84 85 eqtrd φ B M = 1 2 B A
87 86 fveq2d φ B M = 1 2 B A
88 40 13 absmuld φ 1 2 B A = 1 2 B A
89 0red φ 0
90 halfgt0 0 < 1 2
91 90 a1i φ 0 < 1 2
92 89 42 91 ltled φ 0 1 2
93 42 92 absidd φ 1 2 = 1 2
94 93 oveq1d φ 1 2 B A = 1 2 B A
95 87 88 94 3eqtrd φ B M = 1 2 B A
96 95 oveq1d φ B M 2 = 1 2 B A 2
97 40 15 sqmuld φ 1 2 B A 2 = 1 2 2 B A 2
98 96 97 eqtrd φ B M 2 = 1 2 2 B A 2
99 40 16 13 subdird φ 1 2 X B A = 1 2 B A X B A
100 86 31 oveq12d φ B - M - B P = 1 2 B A X B A
101 80 halfcld φ A + B 2
102 4 101 eqeltrd φ M
103 2 102 23 nnncan1d φ B - M - B P = P M
104 99 100 103 3eqtr2rd φ P M = 1 2 X B A
105 104 fveq2d φ P M = 1 2 X B A
106 44 13 absmuld φ 1 2 X B A = 1 2 X B A
107 105 106 eqtrd φ P M = 1 2 X B A
108 107 oveq1d φ P M 2 = 1 2 X B A 2
109 46 15 sqmuld φ 1 2 X B A 2 = 1 2 X 2 B A 2
110 108 109 eqtrd φ P M 2 = 1 2 X 2 B A 2
111 98 110 oveq12d φ B M 2 P M 2 = 1 2 2 B A 2 1 2 X 2 B A 2
112 49 70 111 3eqtr4rd φ B M 2 P M 2 = 1 X X B A 2
113 38 112 eqtr4d φ P A P B = B M 2 P M 2