Inequalities for Triangles and Pointwise Characterizations

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Marsha Tate
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1 Inequalities for Triangles and Pointwise haracterizations Theorem (The Scalene Inequality): If one side of a triangle has greater length than another side, then the angle opposite the longer side has the greatest measure, and conversely. ~ Referring to the diagram, let > and find such that ** and =. Since is interior to p, we have µ(p) > µ(p1) = µ(p). Since µ(p) > µ(p) by the exterior angle inequality, we have µ(p) > µ(p). 1 For the converse, suppose that mp > mp. There are three possibilities for the relationship between and : Either <, =, or >. y what we just proved, we cannot have < or else mp < mp, a contradiction. Moreover, if =, the triangle is isosceles and mp = mp, a contradiction. So, >. orollary 1: If a triangle has an obtuse or right angle, then the side opposite that angle has the greatest length. efinition: triangle is a right triangle if it has a right angle. The side opposite the right angle is called the hypotenuse and the other two sides are called legs. orollary : In a right triangle, the hypotenuse has length greater than that of either leg.

2 Theorem (The Triangle Inequality): In any triangle, the sum of the measures of two sides is greater than that of the third side. More generally: For any three distinct points,, and, + $, with equality if and only if **. ~ ase 1:,, and are not collinear: Given ª, extend to point so that ** and =. Then = + = + since **. Since ª is isosceles, µ(p1) = µ(p), and point is interior to p. So µ(p) = µ(p) + µ(p3) > µ(p) = µ(p1) = µ(p). y the Scalene Inequality, >, so + >. ase :,, and are collinear. Then, by the Ruler Postulate, either **, **, or **. If **, + =; if ** or **, + >. Note: We have shown that if **, + =, and if not **, then + >. This establishes the if and only if statement. 1 3

3 orollary (Median Inequality Not in our Text): Suppose that M M < 1 ( + ) of ª (i.e, M is the midpoint of ). Then. is the median to side ~ Find point such that M is the midpoint of M. Then by SS and PF, =. onsider ª. y the triangle inequality, < + = +. ut so we < +, or M < 1 ( + ). M

4 Theorem (SS Inequality, lligator Theorem or Hinge Theorem): If in ª and ªXYZ we have = XY, = XZ, but µ(p) > µ(px), then > YZ, and conversely, if > ZY, then µ(p) > µ(px). ~ onstruct ray between and with µ(p) = µ(px), and with = XZ =. Then ª ªXYZ by SS, and = YZ by PF. X Y Z s in the figure below, construct the bisector of p. This cuts segment at an interior point E. (Why?) Then pe pe, E = E, and =, so ªE = ªE by SS. Then E = E, and employing the triangle inequality on ªE, and because *E*, = E + E = E + E > = YZ, so > YZ. X E Y Z For the converse, use the same trick as in the Scalene Inequality: Suppose > YZ but µ(p) # µ(px). If µ(p) = µ(px), then = YZ by SS and PF. If µ(p) < µ(px), then the proof we just gave would establish < YZ, a contradiction. So µ(p) > µ(px).

5 Theorem: If l is a line and P is a point not on l, and let F be the foot of the perpendicular from P to l (i.e., the point where the perpendicular to l that contains P intersects l). If R is any point of l, then PR > PF. ~ Immediate from the above corollary that the hypotenuse of a right triangle is longer than either leg. efinition: If l is a line and P is a point not on l, the distance from P to l is the distance from P to the foot F of the perpendicular from P to l. Theorem (Pointwise haracterization of the ngle isector): Let,, and be three noncollinear points and let P be a point in the interior of p. Then P lies on the angle bisector of p if and only if P is equidistant from the sides of the angle, i.e, the lines and. Theorem (Pointwise haracterization of the Perpendicular isector): The set of all points equidistant from each of two points and is the perpendicular bisector of. The proofs of these two theorems are straightforward applications of isosceles triangles and congruence theorems, and make good exercises.

6 Theorem (The ontinuity of istance): Given ray and any point O not on, define a function d(x) for any real x $0 as the distance from O to P on, where x is the distance from to P. That is,. Then d(x) is continuous. d( x) = OP iff x = P O d(0) d(x) x P ~ Pick ε > 0. Let R and S be points of such that R = x and S = y. Let y x = S R = RS < ε. y the Triangle Inequality, OR OS + RS and OS OR + RS. Either way, OS OR RS. Then we have d( y) d( x) = OS OR RS < ε. Thus d(x) is continuous. Note: We will use this theorem later when we prove the Elementary ontinuity of ircles in hapter 10. O d(0) d(y) d(x) S R

Exterior Angle Inequality

Exterior Angle Inequality xterior ngle Inequality efinition: Given Î, the angles p, p, and p are called interior angles of the triangle. ny angle that forms a linear pair with an interior angle is called an exterior angle. In the