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Ch.5 Relationships in Triangles. pp. 234-279. 5-1 Bisectors, Medians, and Altitudes. p.238. perpendicular bisector - in a triangle, a line, segment, or ray that passes through the midpoint of a side and is perpendicular to that side
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Ch.5 Relationships in Triangles pp. 234-279
5-1 Bisectors, Medians, and Altitudes p.238 • perpendicular bisector - in a triangle, a line, segment, or ray that passes through the midpoint of a side and is perpendicular to that side • concurrent lines - three or more lines that intersect at a common point A G is the midpoint of BC. AG is a perpendicular bisector C G B
point of concurrency - the point of intersection of concurrent lines • circumcenter - the point of concurrency of the perpendicular bisectors of a triangle • incenter - the point of concurrency of the angle bisectors of a triangle . . .
(perpendicular bisectors) Thm. 5.1 Any point on the perpendicular bisector of a segment is equidistant from the endpoints of the segment. (example - on board) Equidistant means “the same distance.” Thm. 5.2 Any point equidistant from the endpoints of a segment lies on the perpendicular bisector of the segment. (example - on board) Note that Thm. 5.2 states the point is on the perpendicular bisector. It does not say that any line containing that point is a perpendicular bisector.
Thm. 5.3 Circumcenter Theorem The circumcenter of a triangle is equidistant from the vertices of the triangle. (example - on board) (angle bisectors) Thm. 5.4 Any point on the angle bisector is equidistant from the sides of the angle. (example - on board) Thm. 5.5 Any point equidistant from the sides of an angle lies on the angle bisector. (example - on board)
Thm. 5.6 Incenter Theorem The incenter of a triangle is equidistant from each side of the triangle. (example - on board) • median - in a triangle, a line segment with endpoints that are a vertex of a triangle and the midpoint of the side opposite the vertex • centroid - the point of concurrency of the medians of a triangle Because the median contains the midpoint, it is also a bisector of the side of the triangle.
Thm. 5.7 Centroid Theorem The centroid of a triangle is located two thirds of the distance from a vertex to the midpoint of the side opposite the vertex on a median. (example - on board) • altitude - in a triangle, a segment from a vertex of the triangle to the line containing the opposite side and perpendicular to that side. • orthocenter - the point of concurrency of the altitudes of a triangle NOTE: An orthocenter does not necessarily have to lie in the interior of a triangle. (see example)
5-2 Inequalities and Triangles p.247 Definition of Inequality For any real numbers a and b, a > b if and only if there is a positive number c such that a = b + c. Example:If 6 = 4 + 2, 6 > 4 and 6 > 2. Properties of Inequalities Chart - p. 247 Thm. 5.8 Exterior Angle Inequality Theorem If an angle is an exterior angle of a triangle, then its measure is greater than the measure of either of its corresponding remote interior angles. (example - on board)
NOTE: The symbol for angle () looks similar to the symbol for less than (<), especially when hand written. Be careful to write the symbols correctly in situations where both are used. Thm. 5.9 If one side of a triangle is longer than another side, then the angle opposite the longer side has a greater measure than the angle opposite the shorter side. Thm. 5.10 If one angle of a triangle has a greater measure than another angle, then the side opposite the greater angle is longer than the side opposite the lesser angle.
5-3 Indirect Proof p.255 • indirect reasoning - reasoning that assumes that the conclusion is false and then shows that this assumption leads to a contradiction of the hypothesis or some other accepted fact, like a postulate, theorem, or corollary. Then, since the assumption has been proved false, the conclusion must be true. • indirect proof - in an indirect proof, one assumes that the statement to be proved is false (using indirect reasoning) • proof by contradiction - an indirect proof
Steps for Writing an Indirect Proof Assume that the conclusion is false Show that this assumption leads to a contradiction of the hypothesis, or some other fact, such as a definition, postulate, theorem, or corollary. Point out that because the false conclusion leads to an incorrect statement, the original conclusion must be true.
5-4 The Triangle Inequality p.261 Thm. 5.11 Triangle Inequality Theorem The sum of the lengths of any two sides of a triangle is greater than the length of the third side. (can be used to determine whether three segments can form a triangle) (example - on board) TIP: If the sum of the smallest number and the middle number is greater than the largest number, then each combination of inequalities are true. You need to check the relationships for all three sides, not just two.
Thm. 5.12 The perpendicular segment from a point to a line is the shortest segment from the point to the line. (example on board) Corollary 5.1 The perpendicular segment from a point to a plane is the shortest segment from the point to the plane. (example on board)
5-5 Inequalities Involving Two Triangles p.267 Thm. 5.13 SAS Inequality/Hinge Theorem If two sides of a triangle are congruent to two sides of another triangle and the included angle in one triangle has a greater measure than the included angle in the other, then the third side of the first triangle is longer than the third side of the second triangle. (also called the Hinge Theorem) (example - on board)
Thm. 5.14 SSS Inequality If two sides of a triangle are congruent to two sides of another triangle and the third side in one triangle is longer than the third side in the other, then the angle between the pair of congruent sides in the first triangle is greater than the corresponding angle in the second triangle. (example - on board) Ex.] Use the picture on the board to answer the questions. a.) Write an inequality relating mLDM to mMDN. mLDM > mMDN b.) Find the range of values containing a. - 5/3 < a < 14
K J H L ~ 2.) Alt. Int. ’s are =. Ex.]Given: KL ll JH mJKH + mHKL < mJHK + mKHL JK = HL Prove: JH < KL Statements Reasons 1.) mJKH + mHKL < mJHK + mKHL 1.) Given 2.) mHKL = mJHK 3.) mJKH + mJHK < mJHK + mKHL 3.) Substitution 4.) Subtraction 4.) mJKH < mKHL 5.) JK = HL 5.) Given 6.) Reflexive 6.) HK = HK 7.) SAS Inequality 7.) JH < KL