1 / 16

Chapter 6

Chapter 6. Francisco Tomasino Andy Lachler. Properties of Inequality. If a>b and c≥d , then a+C > B+D If a>b and c>0, then ac> bc and a/c > b/c If a>b and c>0, then ac< bc and a/c <b/c If a>b and b>c , then a>c If a= b+c and c>0, then a>b. Example:.

august-curry
Download Presentation

Chapter 6

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 6 Francisco Tomasino Andy Lachler

  2. Properties of Inequality • If a>b and c≥d, then a+C> B+D • If a>b and c>0, then ac>bc and a/c > b/c • If a>b and c>0, then ac<bc and a/c <b/c • If a>b and b>c , then a>c • If a= b+cand c>0, then a>b

  3. Example: B E Given: AC>BC ; CE >CD Prove: AE > BD C Statements Reasons Given Prop. Of Ineq. SAP Substitution A D • AC>BC; CE>CD • AC+CE > BC+CD • AC+CE=AE; BC+CD=BD • AE>BD

  4. Exterior Angle Inequality Theorem (EAIT) • The measure of an exterior angle of a triangle is grater than the measure of either remote interior angle A A < C and B < C B C

  5. Example: Given: 2 > 1 Prove: 2 > 4 Statements Reasons Given EAIT Substitution VAT Substitution • 2 > 1 • 1 > 3 • 2 > 3 • 3 = 4 • 2 > 4 4 3 1 2

  6. Indirect Proof How to Write an Indirect Proof • Assume temporarily that the conclusion is not true • Reason logically until you reach a contradiction of a known fact • Point out that the temporary assumption must be false, and that the conclusion must then be true

  7. Example: X Given: X = 100° Prove: Y is not a right angle Statements Reasons 1. Given 2. Assumption 3. Sum of angles of a triangle is 180° 4. Substitution 5. Contradiction Y Z • 1. X = 100° • 2. Assume Y is a right angle • 3. X + Y + Z = 180° • 4. 100+90+Z= 180 • 5. Y is not a right angle

  8. Practice Problems V X Given: VY is perpendicular to YZ Prove: angle VXZ is an obtuse angle Y Z t 1 Given: transversal “t” cuts lines “a” and “b”; angle 1 ≠ angle 2 Prove: “a” is not parallel to “b” a 3 b 2

  9. Inequalities for One Triangle • In any triangle, the largest interior angle is opposite the largest side. • The smallest interior angle is opposite the smallest side • The middle-sized interior angle is opposite the middle-sized side R Angle RST > Angle T Because side RT is longer than RS or ST 5 3 T S 4

  10. Theorems and Corollaries Theorem 6-2 : If one side of a triangle is longer than a second side, then the angle opposite the first side is larger than the angle opposite the second side Given: Triangle RST; RT>RS Prove: m of angle RST > m of angle T • By the ruler postulate there is a point Z on segment RT such that RZ = RS. Draw segment SZ • In isosceles triangle RZS, the measure of angle 3 = the measure of angle 2 • Because the measure of angle RST = the measure of angle 1 + the measure of angle 2, you have measure of angle RST > measure of angle 2 • Substitution of measure of angle 3 for the measure of angle 2 yields the measure of RST > measure of angle 3 • Because angle 3 is and exterior angle of triangle ZSR, you have measure of angle 3 > measure of angle T • From the m of RST > m of 3 and m of 3 > m of T, you get m RST > m T R Z 3 2 1 T S

  11. Theorems and Corollaries Theorem 6-3: If one angle of a triangle is larger than a second angle, then the side opposite the first angle is longer than the side opposite the second angle Given: Triangle RST; m of S > m of T Prove: RT > RS R • Assume temporarily that RT is NOT > RS. The either RT = Rs or RT < RS • Case 1: If RT = RS, then m of S = m of T • Case 2: If RT< RS, then m of S < m of T • In either case there is a contradiction of the given fact that m of S > m of T • The assumption that RT is NOT > RS must be false. It follows that RT > RS T S

  12. Theorems and Corollaries Corollary 1: The perpendicular segment from a point to a line is the shortest segment from the point to the line http://www.mathopenref.com/coordpointdist.html Corollary 2: The perpendicular segment from a point to a plane is the shortest segment from the point to the plane

  13. Theorems and Corollaries Theorem 6-4: The sum of the length of any two sides of a triangle is greater than the length of the third side Given: Triangle ABC Prove: (1) AB + BC > AC (2) AB + AC > BC (3) AC + BC > AB • One of the sides. Say segment AB, is the longest side. Then (1) and (2) are true. To prove (3), draw a line, line CZ, through C and perpendicular to line AB. • By Corollary 1 of Theorem 6-3, segment AZ is the shortest segment from A to segment CZ. Also, Segment BZ is the shortest segment from B to segment CZ. C A B Z

  14. Theorems and CorollariesInequalities for Two Triangles SAS Inequality 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. http://www.geogebra.org/en/upload/files/english/nebsary/SasInequality/SAS.html

  15. Theorems and Corollaries SSS Inequality Theorem: 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 http://www.geogebra.org/en/upload/files/english/nebsary/SssInequality/SssFinal.html

  16. Practice Problems Given: Segment BA is congruent to segment ED; Segment BC is congruent to segment EF; m of B > m of E Prove: AC > DF A B C Given: Segment BA is congruent to segment ED; Segment BC is congruent to segment EF; AC > DF Prove: m of B > m of E D E F

More Related