Metamath Proof Explorer


Theorem angpined

Description: If the angle at ABC is _pi , then A is not equal to C. (Contributed by David Moews, 28-Feb-2017)

Ref Expression
Hypotheses angpieqvd.angdef 𝐹 = ( 𝑥 ∈ ( ℂ ∖ { 0 } ) , 𝑦 ∈ ( ℂ ∖ { 0 } ) ↦ ( ℑ ‘ ( log ‘ ( 𝑦 / 𝑥 ) ) ) )
angpieqvd.A ( 𝜑𝐴 ∈ ℂ )
angpieqvd.B ( 𝜑𝐵 ∈ ℂ )
angpieqvd.C ( 𝜑𝐶 ∈ ℂ )
angpieqvd.AneB ( 𝜑𝐴𝐵 )
angpieqvd.BneC ( 𝜑𝐵𝐶 )
Assertion angpined ( 𝜑 → ( ( ( 𝐴𝐵 ) 𝐹 ( 𝐶𝐵 ) ) = π → 𝐴𝐶 ) )

Proof

Step Hyp Ref Expression
1 angpieqvd.angdef 𝐹 = ( 𝑥 ∈ ( ℂ ∖ { 0 } ) , 𝑦 ∈ ( ℂ ∖ { 0 } ) ↦ ( ℑ ‘ ( log ‘ ( 𝑦 / 𝑥 ) ) ) )
2 angpieqvd.A ( 𝜑𝐴 ∈ ℂ )
3 angpieqvd.B ( 𝜑𝐵 ∈ ℂ )
4 angpieqvd.C ( 𝜑𝐶 ∈ ℂ )
5 angpieqvd.AneB ( 𝜑𝐴𝐵 )
6 angpieqvd.BneC ( 𝜑𝐵𝐶 )
7 1 2 3 4 5 6 angpieqvdlem2 ( 𝜑 → ( - ( ( 𝐶𝐵 ) / ( 𝐴𝐵 ) ) ∈ ℝ+ ↔ ( ( 𝐴𝐵 ) 𝐹 ( 𝐶𝐵 ) ) = π ) )
8 1rp 1 ∈ ℝ+
9 1re 1 ∈ ℝ
10 ax-1ne0 1 ≠ 0
11 rpneg ( ( 1 ∈ ℝ ∧ 1 ≠ 0 ) → ( 1 ∈ ℝ+ ↔ ¬ - 1 ∈ ℝ+ ) )
12 9 10 11 mp2an ( 1 ∈ ℝ+ ↔ ¬ - 1 ∈ ℝ+ )
13 8 12 mpbi ¬ - 1 ∈ ℝ+
14 2 3 subcld ( 𝜑 → ( 𝐴𝐵 ) ∈ ℂ )
15 14 adantr ( ( 𝜑𝐶 = 𝐴 ) → ( 𝐴𝐵 ) ∈ ℂ )
16 2 3 5 subne0d ( 𝜑 → ( 𝐴𝐵 ) ≠ 0 )
17 16 adantr ( ( 𝜑𝐶 = 𝐴 ) → ( 𝐴𝐵 ) ≠ 0 )
18 simpr ( ( 𝜑𝐶 = 𝐴 ) → 𝐶 = 𝐴 )
19 18 oveq1d ( ( 𝜑𝐶 = 𝐴 ) → ( 𝐶𝐵 ) = ( 𝐴𝐵 ) )
20 15 17 19 diveq1bd ( ( 𝜑𝐶 = 𝐴 ) → ( ( 𝐶𝐵 ) / ( 𝐴𝐵 ) ) = 1 )
21 20 adantlr ( ( ( 𝜑 ∧ - ( ( 𝐶𝐵 ) / ( 𝐴𝐵 ) ) ∈ ℝ+ ) ∧ 𝐶 = 𝐴 ) → ( ( 𝐶𝐵 ) / ( 𝐴𝐵 ) ) = 1 )
22 21 negeqd ( ( ( 𝜑 ∧ - ( ( 𝐶𝐵 ) / ( 𝐴𝐵 ) ) ∈ ℝ+ ) ∧ 𝐶 = 𝐴 ) → - ( ( 𝐶𝐵 ) / ( 𝐴𝐵 ) ) = - 1 )
23 simplr ( ( ( 𝜑 ∧ - ( ( 𝐶𝐵 ) / ( 𝐴𝐵 ) ) ∈ ℝ+ ) ∧ 𝐶 = 𝐴 ) → - ( ( 𝐶𝐵 ) / ( 𝐴𝐵 ) ) ∈ ℝ+ )
24 22 23 eqeltrrd ( ( ( 𝜑 ∧ - ( ( 𝐶𝐵 ) / ( 𝐴𝐵 ) ) ∈ ℝ+ ) ∧ 𝐶 = 𝐴 ) → - 1 ∈ ℝ+ )
25 24 ex ( ( 𝜑 ∧ - ( ( 𝐶𝐵 ) / ( 𝐴𝐵 ) ) ∈ ℝ+ ) → ( 𝐶 = 𝐴 → - 1 ∈ ℝ+ ) )
26 25 necon3bd ( ( 𝜑 ∧ - ( ( 𝐶𝐵 ) / ( 𝐴𝐵 ) ) ∈ ℝ+ ) → ( ¬ - 1 ∈ ℝ+𝐶𝐴 ) )
27 13 26 mpi ( ( 𝜑 ∧ - ( ( 𝐶𝐵 ) / ( 𝐴𝐵 ) ) ∈ ℝ+ ) → 𝐶𝐴 )
28 27 ex ( 𝜑 → ( - ( ( 𝐶𝐵 ) / ( 𝐴𝐵 ) ) ∈ ℝ+𝐶𝐴 ) )
29 necom ( 𝐶𝐴𝐴𝐶 )
30 28 29 syl6ib ( 𝜑 → ( - ( ( 𝐶𝐵 ) / ( 𝐴𝐵 ) ) ∈ ℝ+𝐴𝐶 ) )
31 7 30 sylbird ( 𝜑 → ( ( ( 𝐴𝐵 ) 𝐹 ( 𝐶𝐵 ) ) = π → 𝐴𝐶 ) )