[alogo] The triangle formed by the bisectors

By an argument similar to one defining the intersection point I of the three internal bisectors (see Bisector0.html ),
two external bisectors and one internal intersect at the same point. Such a point is equidistant from the sides of thetriangle. Thus there exists three excircles with centers at these intersection points Ia, Ib, Ic.
[1] The triangle IaIbIc formed by the external bisectors of ABC has the internal bisectors as altitudes.
[2] The original triangle is the orthic of IaIbIc. Its circumcircle is the Euler cirlce of IaIbIc.
[3] Let Ja, Ja, Jb be the projections on the sides of the incenter I. Let further Jaa, Jab, Jac be the   projections of the excenter Ja on the sides and define similarly Jba, Jbb, etc.. The following relations hold:
  [3.1]  |AJac|=|AJab|=s, where s=(a+b+c)/2 the half-perimeter of the triangle.
  [3.2]  |CJab|=|CJaa|=s-b, |AJc|=|AJb|=s-a.
  [3.3]  |BJa|=|CJaa|=s-b, |JaJaa|=b-c.
  [3.4]  BC and JaJaa have the same middle.
[4]  r-1 = ra-1+rb-1+rc-1.
      Here a, b, c denote the length of BC, CA, AB respectively. Similar formulas hold by replacing the symbols with      a cyclic permutation of a,b,c. Here the symbols {r, ra, rb, rc} denote the radii of the in/excircles respectively.

[0_0] [0_1]
[1_0] [1_1]
[2_0] [2_1]

[1] The orthogonality of inner to outer bisector at C, B, A is trivial.
[2] Trivial consequence of [1] (see Orthic.html ).
[3] All relations are easy consequences of the fact that the tangents to a circle from a point are equal.
[4] Measure the area of ABC in terms of the in/exradii: area(ABC) = s*r. Taking the triangles with bases the     sides of ABC and one vertex at Ib: 2*area(ABC) = (c+a-b)*rb. Hence r/rb = (c+a-b)/(2*s). To prove     the property add the three corresponding formulas, resulting by cyclic permutation of a, b, c.  
[5] The formulas proved here are used in Fundamental_Invariants.html to derive expressions like f.e. the sum of squares     a2 + b2 + c2 in terms of  s, r and R (which are precisely the fundamental invariants of the triangle ABC).

Some further remarks related to the bisectors are contained in Bisector.html .

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