let A, B, C be Point of (TOP-REAL 2); ( angle (B,A,C) <> angle (C,B,A) implies sin (((angle (B,A,C)) - (angle (C,B,A))) / 2) <> 0 )
assume A1:
angle (B,A,C) <> angle (C,B,A)
; sin (((angle (B,A,C)) - (angle (C,B,A))) / 2) <> 0
assume A2:
sin (((angle (B,A,C)) - (angle (C,B,A))) / 2) = 0
; contradiction
then consider i0 being Integer such that
A3:
((angle (B,A,C)) - (angle (C,B,A))) / 2 = PI * i0
by BORSUK_7:7;
set a = (angle (B,A,C)) - (angle (C,B,A));
A4:
( 0 <= angle (B,A,C) & angle (B,A,C) < 2 * PI & 0 <= angle (C,B,A) & angle (C,B,A) < 2 * PI )
by EUCLID11:2;
A5:
0 - (2 * PI) < 0 - (angle (C,B,A))
by EUCLID11:2, XREAL_1:10;
per cases
( i0 = 0 or i0 < 0 or i0 > 0 )
;
suppose
i0 = 0
;
contradictionhence
contradiction
by A3, A1;
verum end; suppose A6:
i0 < 0
;
contradiction
PI in ].0,4.[
by SIN_COS:def 28;
then
0 < PI
by XXREAL_1:4;
then A7:
(angle (B,A,C)) - (angle (C,B,A)) < 0
by A3, A6;
- (2 * PI) < (angle (B,A,C)) - (angle (C,B,A))
by A5, EUCLID11:2, XREAL_1:8;
hence
contradiction
by A2, A7, Th5;
verum end; suppose A8:
i0 > 0
;
contradiction
(
0 < (angle (B,A,C)) - (angle (C,B,A)) &
(angle (B,A,C)) - (angle (C,B,A)) < 2
* PI )
proof
thus
0 < (angle (B,A,C)) - (angle (C,B,A))
by A3, A8, COMPTRIG:5;
(angle (B,A,C)) - (angle (C,B,A)) < 2 * PI
(angle (B,A,C)) - (angle (C,B,A)) < (2 * PI) - 0
by A4, XREAL_1:14;
hence
(angle (B,A,C)) - (angle (C,B,A)) < 2
* PI
;
verum
end; hence
contradiction
by A2, Th4;
verum end; end;