[1] Each is the geometric locus of points equidistant from the sides of the angle it bisects.

[2] The three bisectors intersect at a point I, center of the incircle of the triangle.

[3] The trace of a bisector on the opposite side divides it at ratio equal to the ratio of the sides of the angle it bisects.

[4] The same is true for the trace on the opposite side of the external bisector: A*B/A*C=AB/AC and similarly for the

other traces B* and C* (of the external bisectors). Hence on each side the two traces (internal/external) of the

bisectors are harmonic conjugate (see Harmonic.html ) with respect to the vertices on this side.

[5] The

[6] The external bisectors of the angles cut the opposite sides at three collinear points A*, B*, C*. The line

containingA*,B*,C* is the

[1] A point I on the bisector defines by its projections I

II

[2] Assume I is the intersection point of the bisectors at B and C. Then II

II

[3-4] Let B' be the intersection point of the parallel BB' to the bisector AD. Triangle ABB' is isosceles and by the parallelity of BB'

to AD: BD/DC = B'A/AC = AB/AC. The proof for the external bisector is analogous. The statement on the harmonicity is a

consequence.

[5] AI

[6] Given that A*B/A*C = AB/AC, B*C/B*A=BC/BA, C*A/C*B=CA/CB, apply Menealaus theorem ( Menelaus.html ) etc.. The

harmonicity cited is a sufficient condition for A*, B*, C* in order to be on the trilinear polar of I (see reference below). The

triangle formed by the intersection points of two external and one internal bisector is examined in Bisector1.html .

With the bisectors are related the

the file Incircle.html .

Pedal.html

TrilinearPolar.html

Menelaus.html

Bisector1.html

Incircle.html

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