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Chapter 5: Properties of Triangles. Geometry Fall 2008. 5.1 Perpendiculars and Bisectors. Perpendicular Bisector: a segment, ray, or line that is perpendicular to a segment at its midpoint. Equidistant: the same distance from. Perpendicular Bisector Theorem.
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Chapter 5: Properties of Triangles Geometry Fall 2008
5.1 Perpendiculars and Bisectors • Perpendicular Bisector: a segment, ray, or line that is perpendicular to a segment at its midpoint. • Equidistant: the same distance from
Perpendicular Bisector Theorem • Bisector Theorem: If a point is on the bisector of a segment, then it is equidistant from the endpoints of the segment. If PM bisects AB and is to AB, then AP = BP. P A M B
Converse • If AP = BP, then P is on the bisector of AB.
Angle Bisector Theorem • If a point is on the bisector of an angle then it is equidistant from the two sides of the angle. If m BAD=m CAD then DB = DC. Converse: If DB = DC then m BAD = m CAD B D A C
5.2 Bisectors of a Triangle • Perpendicular bisector of a triangle: line, ray, or segment that is perpendicular to a side of the triangle at the midpoint of the side.
Concurrent Lines • When 3 or more lines intersect in the same point they are called concurrent lines. • Then intersection is called the concurrency. • The 3 perpendicular bisectors of a triangle are concurrent.
Where do the perp. bisectors of an acute, right, or obtuse triangle meet? • Inside On Outside
Circumcenter • The intersection of the three perpendicular bisectors is called the circumcenter. • The circumcenter is equidistant from each of the vertices.
Angle Bisector of a Triangle • Angle bisector of a triangle: a line, segment, or ray that bisects and angle of the triangle
Incenter • The 3 angle bisectors are concurrent. • The intersection of the 3 angle bisectors is called the incenter.
Incenter • The incenter is always inside the triangle and is equidistant from the sides of the triangle. • GB = GD = GF D F G B
Is P on the bisector of A? 4 P A 4
5.2 examples Find the length of DA. Find the length of AB. Why is ADF = BDE? B D E 2 A C F
Examples • V is the incenter of XWZ. • Length of VS: • m VZX = W S T 3 V 20 Z X Y
Find ID C 8 10 D I A B
Find BD C D 15 A B 12 12
Homework p. 268 # 8-13, 16-26 p. 275 # 10-17
5.3 Medians and Altitudes of a Triangle • Median of a Triangle: a segment whose endpoints are a vertex of the triangle and the midpoint of the opposite side.
Centroid • The three medians are concurrent. • The intersection of the three medians is called the centroid of the triangle.
Concurrency of Medians of a triangle • The medians of a triangle intersect at a point that is two thirds of the distance from each vertex to the midpoint of the opposite side. • If P is the centroid of ABC, then AP = 2/3 AD, BP = 2/3 BF, and CP = 2/3 CE. B D C E P F A
Altitudes of a Triangle • The altitude of a triangle is the perpendicular segment from a vertex to the opposite side or to the line that contains the opposite side. • The intersection is called the orthocenter.
BAE = EAC and BF = FC • Median: • Altitude: • Bisector: • Bisector: • a) AD • b) AE • c) AF • d) GF A G C B D E F
C is the centroid of XYZ. YI = 9.6 X • CK = • XK = • YC = • KZ = • JZ = • Perimeter of : 8 J I C Y Z K 5
5.4 Midsegment Theorem • A midsegment of a triangle is a segment that connects the midpoints of two sides of a triangle.
Midsegment Theorem • The segment connecting the midpoints of two sides of a triangle is parallel to the third side and is half as long.
Example 1 • D, E, and F are the midpoints of the sides. • DE ll _____ • FE ll _____ • If AB = 14, then EF = _____ • If DE = 6, then AC = _____ A D F B C E
Example 2 • R, S, and T are the midpoints of the sides. • RK = 3, KS = 4, and JK KL • RS = _____ • JK = _____ • RT = _____ • Perimeter = _____ J T R 3 K 4 S L
Example 3 • If YZ = 3x + 1, and MN = 10x – 6, then YZ = ______ M X Z O N Y
Example 4 • If m MON = 48 , then m MZX = _____ M X Z O N Y
Example 5 • Find the midpoint of JK. • Find the length of NK. • Find the length of NP. • Find the coordinates of point P.
Homework for 5.3-5.4 • P. 282 #8-11, 17-20 • P. 290 # 12-18, 28-29
5.5 Inequalities in One Triangle • If side 1 of a triangle is larger than side 2, then the angle opposite side 1 is larger than the angle opposite side 2. B Since BC > AB, A > C 5 3 C A
Converse • If angle 1 is larger than angle 2, then the side opposite angle 1 is longer than the side opposite angle 2. B Since A > C, BC > AB 60 40 C A
Exterior Angle Inequality • The measure of an exterior angle of a triangle is greater than the measure of either of the two nonadjacent interior angles. • m 1 > m A and m 1 > m B A 1 B C
Triangle Inequality • The sum of the lengths of any 2 sides of a triangle is greater than the length of the 3rd side • AB + BC > AC • AC + BC > AB • AB + AC > BC A C B
Example 1 • Name the shortest and longest sides. Q 35 65 P R
Example 2 • Name the smallest and largest angles. Z 4 X 6 8 Y
Example 3 • List the sides in order from shortest to longest. B C 30 95 A
Example 4 • Find the possible measures for XY in XYZ. • XZ = 2 and YZ = 3
Example 5 • Determine whether the given lengths could represent the lengths of the sides of a triangle: • A) { 5, 5, 8} • B) { 5, 6, 7} • C) { 1, 2, 5} • D) { 7, 3, 9}
Homework 5.5 • Pg. 298 # 6-11, 14-19, 24-25
5.6 Indirect Proof and Inequalities in Two Triangles • Indirect Proof: a proof in which you prove that a statement is true by first assuming that its opposite is true.
Writing an indirect proof • Assume that the negation of the conclusion (what you are trying to prove) is true. • Show that the assumption leads to a contradiction of known facts or of the given information. • Conclude that since the assumption is false the original conclusion is true.
Example 1 • Given: ABC • Prove: ABC does not have more than one obtuse angle. • Begin by assuming that ABC does have more than one obtuse angle. • m A > 90 and m B > 90 • m A + m B > 180
Example 1 • You know that the sum of the measures of all three angles is 180 . • m A + m B + m C = 180 • m A + m B = 180 - m C • You can substitute 180 - m C for m A + m B in m A + m B > 180 . • 180 - m C > 180 • 0 > m C
Example 1 • The last statement is “not possible.” Angle measure in triangles cannot be negative. • So you can conclude that the original assumption must be false. That is, ABC cannot have more than one obtuse angle.