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

[1] The triangle I

[2] The original triangle is the

[3] Let J

[3.1] |AJ

[3.2] |CJ

[3.3] |BJ

[3.4] BC and J

[4] r

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, r

[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 I

[5] The formulas proved here are used in Fundamental_Invariants.html to derive expressions like f.e. the sum of squares a

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

Orthic.html

Fundamental_Invariants.html

Bisector.html

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