
theorem
  for A,B,C,P being Point of TOP-REAL 2 st A,B,C is_a_triangle &
  A,B,P is_a_triangle &
  angle (C,B,A) < PI &
  angle (A,P,B) < PI &
  angle (P,B,A) = angle (C,B,A) / 3 &
  angle (B,A,P) = angle (B,A,C) / 3 &
  sin (PI/3 - angle(A,C,B)/3)<>0 holds
  |.A-P.| =
  (- the_diameter_of_the_circumcircle(C,B,A) * 4 * sin (angle(A,C,B)/3) *
  sin (PI/3 + angle(A,C,B)/3) * sin (angle(C,B,A) /3))
  proof
    let A,B,C,P be Point of TOP-REAL 2;
    assume that
A1: A,B,C is_a_triangle and
A2: A,B,P is_a_triangle and
A3: angle(C,B,A) < PI and
A4: angle (A,P,B) < PI and
A5: angle (P,B,A) = angle (C,B,A) /3 and
A6: angle (B,A,P) = angle (B,A,C) /3 and
A7: sin (PI/3 - angle(A,C,B)/3)<>0;
A8: angle (C,B,A)/3 + angle (B,A,C)/3 + angle (A,C,B)/3 = PI/3
    by A1,A3,Lm12;
    A,B,P are_mutually_distinct by A2,EUCLID_6:20; then
A9: |.A-P.| * sin (2*PI/3 +angle(A,C,B)/3) =
    |.A-B.| * sin (angle(C,B,A) /3) by A5,A6,A4,A8,Thm35;
    sin angle (A,C,B) = sin (3 * (angle(A,C,B)/3))
    .= 4 * sin (angle(A,C,B)/3) * sin (PI/3 + angle(A,C,B)/3) *
    sin (PI/3 - angle(A,C,B)/3) by Thm18; then
A10: |.A-P.| * sin (2*PI/3 + (angle(A,C,B)/3)) =
    the_diameter_of_the_circumcircle(A,B,C)* (4 * sin (angle(A,C,B)/3) *
    sin (PI/3 + angle(A,C,B)/3) * sin (PI/3 - angle(A,C,B)/3)) *
    sin (angle(C,B,A) /3) by A9,A1,Thm34;
    |.A-P.| * sin (PI/3 - angle(A,C,B)/3) =
    the_diameter_of_the_circumcircle(A,B,C)* (4 * sin (angle(A,C,B)/3) *
    sin (PI/3 + angle(A,C,B)/3) * sin (PI/3 - angle(A,C,B)/3)) *
    sin (angle(C,B,A) /3) by A10,Thm7;
    then |.A-P.| =
    the_diameter_of_the_circumcircle(A,B,C) * 4 * sin (angle(A,C,B)/3) *
    sin (PI/3 + angle(A,C,B)/3) * sin (angle(C,B,A) /3) *
    sin (PI/3 - angle(A,C,B)/3) / sin (PI/3 - angle(A,C,B)/3)
    by A7,XCMPLX_1:89; then
A11: |.A-P.| =
    (the_diameter_of_the_circumcircle(A,B,C) * 4 * sin (angle(A,C,B)/3) *
    sin (PI/3 + angle(A,C,B)/3) * sin (angle(C,B,A) /3)) *
    (sin (PI/3 - angle(A,C,B)/3) / sin (PI/3 - angle(A,C,B)/3))
    by XCMPLX_1:74;
    sin (PI/3 - angle(A,C,B)/3) / sin (PI/3 - angle(A,C,B)/3) = 1
    by A7,XCMPLX_1:60;
    then |.A-P.| =
    (- the_diameter_of_the_circumcircle(C,B,A)) * 4 * sin (angle(A,C,B)/3) *
    sin (PI/3 + angle(A,C,B)/3) * sin (angle(C,B,A) /3)
    by A11,A1,Thm33;
    hence thesis;
  end;
