The bisectors of the outer angles at the vertices B and C of the triangle ABC intersect at the point O. Prove that the ray AO is the bisector of the angle BAC.
From point O we construct perpendiculars OK, OH, OK to straight lines AB, BC and AC.
Triangles ОВК and ОВН are rectangular and equal, since they have a common hypotenuse of ОВ, and the angle ОВН = ОВК, since BO is a bisector, then OK = ОН.
Similarly, the triangle OCH = OCM, and then OM = OH.
Consequently, OK = OH = OK, which means that through the points K, H, C you can draw a circle with center at point O.
Triangles AKO and AMO are rectangular, in which OK = OM as the radii of the circle, AO is the common hypotenuse, then the triangles are equal in leg and hypotenuse. Consequently, the angle KAO = MAO, and the AO is the bisector of the angle BKM and BАС, which was required to be proved.
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