AMSCO'S Geometry. New York - Rye High School

364 Parallel LinesConcurrence of Angle Bisectors of a TriangleIn earlier chapters, we saw that the perpendicular bisectors of the sides of a triangleintersect in a point and that the altitudes of a triangle intersect in a point.Now we can prove that the angle bisectors of a triangle intersect in a point.Theorem 9.15The angle bisectors of a triangle are concurrent.Given ABC with AL the bisector of A, BMthe bisector of B, and CN the bisectorof C.CProve AL, BM, and CN intersect in a point, P.MLProof Let P be the point at which AL and BMPintersect. If a point lies on the bisector ofan angle, then it is equidistant from theANBsides of the angle. Therefore, P is equidistantfrom AC and AB because it lies on the bisector of A, and P is equidistantfrom AB and BC because it lies on the bisector of B. Therefore, P isequidistant from AC, AB, and BC. If a point is equidistant from the sides ofan angle, then it lies on the bisector of the angle. Since P is equidistant fromAC and BC, then it lies of the bisector of C. Therefore, the three anglebisectors of ABC intersect at a point, P.The point where the angle bisectors of a triangle are concurrent is called theincenter.EXAMPLE 1Given: ABC, AB ⊥ BD, AB DC, andAD BC.CDProve: DAB BCDProofWe can show that ADB and CBD areright triangles and use HL to prove themcongruent.AB