[1] The bisector AI intersects the circumcircle k of triangle ABC at F, the middle of arc BC.

[2] The middles of the sides of the triangle of excenters I

[3] |FB|=|FC|=|FI| and similar equalities for the other points analogous to F.

[4] angle(AFM) = angle(B)-angle(C).

[1] That F is the middle of arc BC follows from the equality of angles ABF=FAC. Besides quadrilateral IBEC is cyclic, since IBE and ICE are right angles.

[2] Since I

[3] Follows from the previous two statements.

[4] Consider the symmetric B' of B w.r. to the bisector AI: angle(AFM)=angle(B'BC)=B-C.

With the bisectors of a triangle are connected the following two important theorems:

a)

b)

With the bisectors are connected also some difficult (or impossible) to resolve triangle constructions. An interesting case is examined in the file TriangleBisectorConstruction.html .

Besides, another important related subject that adds much structure into the geometry of the triangle is the one on the

Bisector1.html

BisectorRectangle.html

Euler.html

TriangleBisectorConstruction.html

TriangleBisectors.html

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