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
Let’s carry out additional constructions. Let’s construct the perpendiculars OK, OM and OH to the sides AB, AC BC of the triangle ABC.
In right-angled triangles ОBК and ОВН, the hypotenuse ОB is common, and the angle ОBK = ОBН since ОВ is the bisector of the angle СВН, then the triangles ОВК and ОВН are equal in hypotenuse and acute angle.
Similarly, right-angled triangles OCH and OCM are equal in the hypotenuse OH and the angles OCH and OCM.
Then the segment OM = OH = OM, and therefore point O is the center of the circle, and points K, H, M are points of tangency.
Then the segments AM and AK are tangents to the circle drawn from one point, and then, by the property of tangents, OA is the bisector of the angle BAC, which was required to be proved.
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