let A, B, C be Point of (TOP-REAL 2); ( A <> B & |.(C - A).| = |.(C - B).| implies C in the_perpendicular_bisector (A,B) )
assume that
A1:
A <> B
and
A2:
|.(C - A).| = |.(C - B).|
; C in the_perpendicular_bisector (A,B)
assume A3:
not C in the_perpendicular_bisector (A,B)
; contradiction
consider L1, L2 being Element of line_of_REAL 2 such that
the_perpendicular_bisector (A,B) = L2
and
A4:
L1 = Line (A,B)
and
A5:
L1 _|_ L2
and
L1 /\ L2 = {(the_midpoint_of_the_segment (A,B))}
by A1, Def2;
reconsider rA = A, rB = B, rC = C as Element of REAL 2 by EUCLID:22;
L1 is being_line
by A5, EUCLIDLP:67;
then consider L3 being Element of line_of_REAL 2 such that
A6:
rC in L3
and
A7:
L1 _|_ L3
by Th35;
consider x being Element of REAL 2 such that
A8:
L1 /\ L3 = {x}
by A7, Th33;
reconsider D = x as Point of (TOP-REAL 2) by EUCLID:22;
A9:
D in L1 /\ L3
by A8, TARSKI:def 1;
A10:
now ( not A = D & not C = D & not B = D )assume
(
A = D or
C = D or
B = D )
;
contradictionper cases then
( A = D or C = D or B = D )
;
suppose
C = D
;
contradictionthen
C in L1 /\ L3
by A8, TARSKI:def 1;
then A15:
C in Line (
A,
B)
by A4, XBOOLE_0:def 4;
A16:
|.(A - C).| =
|.(C - B).|
by A2, EUCLID_6:43
.=
|.(B - C).|
by EUCLID_6:43
;
then
C in LSeg (
A,
B)
by A15, Th42;
then
C = the_midpoint_of_the_segment (
A,
B)
by A16, Th28;
hence
contradiction
by A3, A1, Th43;
verum end; end; end;
A20:
( L1 _|_ L3 & D in L1 /\ L3 & A in L1 & C in L3 & A <> D & C <> D )
by A7, A8, TARSKI:def 1, A4, RLTOPSP1:72, A6, A10;
A21:
( L1 _|_ L3 & D in L1 /\ L3 & B in L1 & C in L3 & B <> D & C <> D )
by A7, A8, TARSKI:def 1, A4, RLTOPSP1:72, A6, A10;
set a1 = |.(A - D).|;
set b1 = |.(C - D).|;
set c1 = |.(C - A).|;
A22: |.(C - A).| ^2 =
((|.(A - D).| ^2) + (|.(C - D).| ^2)) - (((2 * |.(A - D).|) * |.(C - D).|) * (cos (angle (A,D,C))))
by EUCLID_6:7
.=
((|.(A - D).| ^2) + (|.(C - D).| ^2)) - (((2 * |.(A - D).|) * |.(C - D).|) * 0)
by A20, SIN_COS:77, Th38
.=
(|.(A - D).| ^2) + (|.(C - D).| ^2)
;
set a2 = |.(B - D).|;
set b2 = |.(C - D).|;
set c2 = |.(C - B).|;
|.(C - B).| ^2 =
((|.(B - D).| ^2) + (|.(C - D).| ^2)) - (((2 * |.(B - D).|) * |.(C - D).|) * (cos (angle (B,D,C))))
by EUCLID_6:7
.=
((|.(B - D).| ^2) + (|.(C - D).| ^2)) - (((2 * |.(B - D).|) * |.(C - D).|) * 0)
by A21, Th38, SIN_COS:77
.=
(|.(B - D).| ^2) + (|.(C - D).| ^2)
;
then A23:
(|.(A - D).| ^2) + (|.(C - D).| ^2) = (|.(B - D).| ^2) + (|.(C - D).| ^2)
by A2, A22;
then
( D in L1 & |.(A - D).| = |.(B - D).| )
by SQUARE_1:40, A9, A23, XBOOLE_0:def 4;
then
( D in LSeg (A,B) & |.(A - D).| = |.(B - D).| )
by A4, Th42;
then
( D = the_midpoint_of_the_segment (A,B) & D in L3 )
by Th28, A9, XBOOLE_0:def 4;
hence
contradiction
by A3, A6, A1, A4, A7, Th44; verum