AMSCO'S Geometry. New York - Rye High School

An Inequality Involving an Exterior Angle of a Triangle 277At vertex A, we can also draw AF h, theopposite ray of AB h, to form DAF, anotherexterior angle of the polygon at vertex A.Ateach vertex of a polygon, two exterior anglescan be drawn. Each of these exterior anglesforms a linear pair with the interior angle atA, and the angles in a linear pair are supplementary.The two exterior angles at A arecongruent angles because they are verticalangles. Either can be drawn as the exteriorangle at A.FEADCBExterior Angles of a TriangleDACFBEAn exterior angle of a triangle is formed outside the triangle by extending a sideof the triangle.The figure to the left shows ABC whose three interior angles are CAB,ABC, and BCA. By extending each side of ABC, three exterior angles areformed, namely, DAC, EBA, and FCB.For each exterior angle, there is an adjacent interior angle and two remoteor nonadjacent interior angles. For ABC, these angles are as follows:Adjacent NonadjacentVertex Exterior Angle Interior Angle Interior AnglesA DAC CAB ABC and BCAB EBA ABC CAB and BCAC FCB BCA CAB and ABCWith these facts in mind, we are now ready to prove another theorem aboutinequalities in geometry called the exterior angle inequality theorem.Theorem 7.2The measure of an exterior angle of a triangle is greater than the measure ofeither nonadjacent interior angle.Given ABC with exterior BCD at vertex C;A and B are nonadjacent interiorangles with respect to BCD.ProvemBCD mBABMCED