Chapter 7. Geometric Inequalities

4. Let m S, then 3 2 m R. Since the angles are supplementary: 3 2580 4568 542 Therefore, m S 42 and m R 38. Part IV 5. Statements Reasons. ABC is not scalene.. Assumption. 2. ABC has at least 2. Definition of a one pair of congruent scalene triangle. sides. 3. A B or 3. Isosceles triangle B C or A C 4. m A m B or 4. Definition of m B m C or m A m C 5. m A m B, 5. Given m B m C, m A m C 6. ABC is scalene. 6. Contradiction. 6. Statements Reasons. T 3, 3 ( ABC). Given. A B C 2. A B AB, 2. In a translation, B C BC distance is preserved. 3. XY A B, 3. Given. YZ B C 4. XY AB, 4. Transitive property YZ BC of equality. 5. XY>AB, 5. Definition of YZ>BC congruent segments. 6. Y B 6. Given. 7. ABC XYZ 7. SAS. Chapter 7. Geometric Inequalities 7- Basic Inequality Postulates (pages 266 267). No. The refleive and symmetric properties do not hold for inequality. 2. No. By the trichotomy postulate, it is possible that AB BC. 3. a. True 4. a. False b. The shortest distance between two points is the length of the line segment joining these two points. 5. a. True 6. a. True b. Definition of a linear pair of angles 7. a. False b. Definition of straight angle 8. a. True b. Trichotomy postulate 9. a. True b. Transitive property of inequality 0. a. True b. Transitive property of inequality. a. True b. Substitution postulate 2. a. True b. Substitution postulate 3. Statements Reasons. AC>BC. Given. 2. A CBA 2. Isosceles triangle 3. m A m CBA 3. Definition of 4. m CBD m CBA 4. Given. 5. m CBD m A 5. Substitution 4. Statements Reasons. PQRS. Given. 2. PR PQ 2. A whole is greater than any of its parts. 3. PQ RS 3. Given. 4. PR RS 4. Substitution 276

5. Statements Reasons. LM>NM. Given. 2. LM NM 2. Definition of congruent segments. 3. KLM 3. Given. 4. KM LM 4. A whole is greater than any of its parts. 5. KM NM 5. Substitution 6. Statements Reasons. KM KN,. Given. KN NM 2. KM NM 2. Transitive property of inequality. 3. NM NL 3. Given. 4. KM NL 4. Substitution 7-2 Inequality Postulates Involving Addition and Subtraction (page 269). No. It is possible for the differences to be same order or equal. 2. Subtracting unequal quantities is the same as adding the opposites of the unequal quantities. To find the opposites, we multiply the inequality by, which changes the order of the inequality. Then, if unequal quantities are added to equal quantities, the sums are same order. 3. If equal quantities (8 8) are added to unequal quantities (0 7), then the sums are unequal in the same order (8 5). 4. If equal quantities ( ) are added to unequal quantities (4 4), then the sums are same order (5 25). 5. If equal quantities (3 3) are subtracted from unequal quantities ( + 3 2), then the differences are same order ( 9). 6. If equal quantities (5 5) are added to unequal quantities (y 5 5), then the sums are unequal in the same order (y 0). 7. If unequal quantities (8 6) are added to unequal quantities in the same order (5 3), then the sums are same order (3 9). 8. If equal quantities (2 2) are subtracted from unequal quantities (7 2), then the differences are same order (5 0). 9. If equal quantities ( ) are subtracted from unequal quantities (y 8), then the differences are same order (y 7). 0. If equal quantities (a b) are subtracted from unequal quantities (80 90), then the differences are same order (80 a 90 b).. Statements Reasons. AB AD,. Given. BC DE 2. AB BC AD DE 2. If equal quantities are added to then the sums are 3. AC AB + BC, 3. Partition AE AD DE 4. AC AE 4. Substitution 2. Statements Reasons. AE BD,. Given. AF BF 2. AE AF BD BF 2. If equal quantities are subtracted from unequal quantities, then the differences are 3. AE AF FE, 3. Partition BD BF FD 4. FE FD 4. Substitution 3. Statements Reasons. AE BE. Given. 2. AE>BE 2. Definition of congruent segments. 3. EAB EBA 3. Isosceles triangle 4. m EAB m EBA 4. Definition of 5. m DAC m DBC 5. Given. 6. m DAC m EAB 6. If equal quantities m DBC m EBA are added to then their sums are (Cont.) 277

Statements Reasons 7. m DAB m DAC 7. Partition m EAB, m CBA m DBC m EBA 8. m DAB m CBA 8. Substitution 4. Yes. If equal weights are added to unequal weights (Blake weighs more than Caleb), then their new weights are same order (Blake still weighs more than Caleb). 5. Not necessarily. If unequal weights are subtracted from unequal weights (Blake weighs more than Andre), then their new weights may be same order (Blake weighs more than Andre), equal in different orders (Andre weighs more than Blake), or equal (Blake and Andre weigh the same). 7-3 Inequality Postulates Involving Multiplication and Division (page 272). No. It is only true for positive values of a. Otherwise, if unequal quantities ( 2) are multiplied by negative equal quantities, then the products are opposite order (a 2a). If a 0, then a 2a. 2. No. It is possible that ac 0 and bd 0. Then a b. It is also possible that ac 0 and bd 0 or ac 0 and bd 0. Then ac bc. 3. If unequal quantities (8 7) are multiplied by positive equal quantities (3 3), then the products are same order (24 2). 4. If unequal quantities (30 35) are divided by negative equal quantities ( 5 5), then the quotients are opposite order ( 7 6). 5. If unequal quantities (8 6) are divided by positive equal quantities (2 2), then the quotients are same order (4 3). 6. If unequal quantities (3 5) are divided by positive equal quantities (3 3), then the quotients are same order ( 5). 7. If unequal quantities A 2.24B are multiplied by negative equal quantities ( 2 2), then the products are opposite order (8 ). A y 6, 3B 8. If unequal quantities are multiplied by positive equal quantities (6 6), then the products are same order (y 8). 9. Always true; if unequal quantities are multiplied by positive equal quantities, then the products are 0. Always true; if equal quantities are added to then the sums are unequal in the. Never true; let a 5, b 4, and c. Then 5 4 or 4 3. 2. Always true; if equal quantities are subtracted from then the differences are 3. Sometimes true; let a 5, b 3, and c 2. Then 5 3 3 2 or 2. Let a 5, b 4, and c 2. Then 5 4 4 2 or 2. 4. Never true; let a 5, b 3, and c 2. Then 2 5, 3 2. 5. Always true; if unequal quantities are divided by positive equal quantities, then the quotients are 6. Never true; let a 3, b 2, and c. Then 3 2. 7. Always true; transitive property 8. Statements Reasons. BD BE. Given. 2. D is the midpoint 2. Given. of BA, E is the midpoint of BC. 3. BA 2BD, 3. Definition of BC 2BE midpoint. 4. 2BD 2BE 4. If unequal quantities are multiplied by positive equal quantities, then the products are unequal in the 5. BA BC 5. Substitution 9. Statements Reasons. m DBA. Given m CAB 2. 2m DBA 2. If unequal quantities 2m CAB are multiplied by positive equal quantities, then the products are unequal in the 278

Statements Reasons 3. m CBA 2m DBA, 3. Given. m DAB m CAB 4. m CBA m DAB 4. Substitution 20. Statements Reasons. AB AD. Given. 2. 2 AB. 2 AD 2. If unequal quantities are multiplied by positive equal quantities, then the products are 3. AE5 2 AB, AF5 2 AD 3. Given. 4. AE AF 4. Substitution 2. Statements Reasons. m CAB m CBA. Given. 2. m CAB m CAD 2. Partition m DAB, m CBA m CBE m EBA 3. m CAD m DAB, 3. Definition of angle m CBE m EBA bisector. 4. m CAB 2m DAB, 4. Substitution m CBA 2m EBA 5. 2m DAB 2m EBA 5. Substitution 6. m DAB m EBA 6. If unequal quantities are divided by positive equal quantities, then the quotients are 7-4 An Inequality Involving the Lengths of the Sides of a Triangle (pages 275 276). If s is not the longest side, then 2 is the longest side. By the triangle inequality theorem, 2 s + 7, so s 5. 2. a. No. This may not be true for negative values. Eample: 5 4 3, but 5 3 ( 4) or 5 7. b. Yes. If a, b, and c are the lengths of the sides of a triangle, then they are positive. By the triangle inequality theorem, a b c. 3. Yes 4. No 5. Yes 6. No 7. Yes 8. No 9. Yes 0. Yes. 2 s 6 2. 9 s 43 3. 0 s 3 4. 2.9 s 22. 5. Since 3 = + 2, by the triangle inequality theorem,, 2, and 3 cannot be the lengths of the sides of a triangle. 6. a 4; solve the inequalities given by the triangle inequality theorem: a,a 2 a 2 2 0,a a 2 2,a a 2 24,a a 2,a a 2 2 4,a 7. Statements Reasons. ABCD is a. Given. quadrilateral. 2. AD AC CD, 2. Triangle inequality AC AB BC 3. AC CD 3. If equal quantities AB BC CD are added to then the sums are 4. AD AB BC CD 4. Transitive property 8. Statements Reasons. AB AD BD. Triangle inequality 2. BC BD DC 2. Partition 3. AD DC 3. Given. 4. BC BD AD 4. Substitution 5. AB BC 5. Substitution 9. Statements Reasons. PZ PQ QZ. Triangle inequality 2. PY PZ 2. If equal quantities PY PQ QZ are added to then the sums are (Cont.) 279

Statements Reasons 3. YQ XY XQ 3. Triangle inequality 4. YQ PY PQ 4. Partition 5. PY PQ XY XQ 5. Substitution 6. PY PQ QZ 6. If equal quantities XY XQ QZ are added to then the sums are 7. XZ XQ QZ 7. Partition 8. PY + PQ QZ 8. Substitution XY XZ 9. PY PZ XY XZ 9. Transitive property Hands-On Activity a., 6 in.; 2, 5 in.; 2, 6 in.; 3, 4 in.; 3, 5 in.; 3, 6 in.; 4, 4 in.; 4, 5 in.; 5, 5 in.; 5, 6 in.; 6, 6 in. b. {, 6, 6}, {2, 5, 6}, {2, 6, 6}, {3, 4, 6}, {3, 5, 6}, {3, 6, 6}, {4, 4, 6}, {4, 5, 6}, {5, 5, 6}, {5, 6, 6}, {6, 6, 6} c. {,, 6}, {, 2, 6}, {, 3, 6}, {, 4, 6}, {, 5, 6}, {2, 2, 6}, {2, 3, 6}, {2, 4, 6}, {3, 3, 6} d. In part b, the sum of the two shortest sides is greater than 6. In part c, the sum is less than or equal to 6. 7-5 An Inequality Involving an Eterior Angle of a Triangle (pages 280 28). Yes. By the eterior angle inequality theorem, the eterior angles that are not adjacent to the right angle have measures greater than the right angle, that is, greater than 90. Therefore, they are obtuse. 2. Yes. The eterior angle adjacent to the right angle forms a linear pair with the right angle. Therefore, it also measures 90 and is a right angle. 3. a. TRP b. T, S 4. a. True b. Definition of median 5. a. True 6. a. True b. Eterior angle inequality theorem 7. a. True 8. a. True b. Eterior angle inequality theorem 9. a. True b. Eterior angle inequality theorem 0. a. True. False 2. a. True 3. False 4. Statements Reasons. Let N be the midpoint. Every line segment of AC. has one and only one midpoint. 2. Draw BN h, etending 2. Two points the ray through N to determine a line. A point G so that line segment can BN>NG. be etended to any length. 3. Draw GC. 3. Two points determine a line. 4. Etend BC through C 4. A line segment can to point F. be etended to any length. 5. m ACF m ACG 5. Partition m FCG 6. AN>NC 6. Definition of midpoint. 7. ANB GNC 7. Vertical angles are 8. ANB CNG 8. SAS (steps 2, 6, 7) 9. A ACG 9. Corresponding parts of congruent triangles are 0. m ACF m ACG 0. A whole is greater than any of its parts.. m ACF m A. Substitution 2. ACF BCD 2. Vertical angles are 3. m BCD m A 3. Substitution 5. Statements Reasons. ABD DBE. Given. ABE, ABE EBC ABC 2. m ABD m ABE, 2. A whole is greater m ABE m ABC than any of its parts. 3. m ABD m ABC 3. Transitive property 280

6. Statements Reasons. DE EF. Given. 2. DE>EF 2. Definition of congruent segments. 3. EFD EDF 3. Isosceles triangle 4. m EFD m EDF 4. Definition of 5. m EFG m EDF 5. Eterior angle inequality 6. m EFG m EFD 6. Substitution 7. Statements Reasons. ABC with. Given. m C 90. 2. Etend AC through 2. A line segment may C to point D, so that be etended to any ACD is formed. length. 3. Eterior angle BCD 3. Definition of an and interior angle eterior angle. BCA form a linear pair. 4. m BCD m MCA 4. If two angles form a 80 linear pair, then they are supplementary. 5. m BCD 90 80 5. Substitution 6. m BCD 90 6. Subtraction 7. 0 m A 90 7. Eterior angle inequality 8. A is acute. 8. Definition of an acute angle. 8. Statements Reasons. m RMP m RTM,. Eterior angle m RTM m SRT inequality 2. m RMP m SRT 2. Transitive property 9. Statements Reasons. FC FD. Given. 2. FC>FD 2. Definition of congruent segments. 3. DCF CDF 3. Isosceles triangle 4. EDF BCF 4. If two angles are congruent, then their supplements are 5. m EDF m BCF 5. Definition of 6. m ABF m BCF 6. Eterior angle inequality 7. m ABF m EDF 7. Substitution 7-6 Inequalities Involving Sides and Angles of a Triangles (pages 284 285). a. If the measures of two angles of a triangle are equal, then the lengths of sides opposite these angles are equal. b. Yes 2. a. If two angles of a triangle are congruent, then sides opposite these angles are b. The converse statement and the contrapositive statement both handle the same relationship between the angles and the sides opposite the angle in a triangle. However, the converse deals with congruence and the contrapositive deals with equality. 3. B 4. DF 5. AC 6. AB 7. B 8. C 9. RS 0. CD.BC.BD. GH,FG,FH 2. RT RS ST 3. Statements Reasons. m ABC m CBD. Given. 2. m CBD m CAB 2. Eterior angle inequality 3. m ABC m CAB 3. Transitive property 4. AC BC 4. If the measures of two angles of a triangle are unequal, then the lengths of the sides of opposite these angles are unequal and the longer side is opposite the larger angle. 4. a. The measure of the eterior angle adjacent to C measures 90 degrees. By the eterior angle inequality theorem, the nonadjacent interior angles, A and B, have measures less than 90 degrees. By definition, A and B are acute. 28

b. C is the largest angle, so hypotenuse AB, which is opposite C, is the longest side. 5. The eterior angle adjacent to the obtuse angle forms a linear pair with the obtuse angle. Since the measure of the obtuse angle is greater than 90 degrees, the measure of the eterior angle is less than 90 degrees. By the eterior angle inequality theorem, the measures of the remote interior angles are also less than 90 degrees. Therefore, the two angles are acute. Review Eercises (pages 286 287). A whole is greater than any of its parts. 2. Transitive property of inequality 3. Eterior angle inequality theorem 4. If the measures of two angles of a triangle are unequal, then the lengths of the sides opposite these angles are unequal and the longer side lies opposite the larger angle. 5. If equal quantities are added to unequal quantities, then the sums are same order. 6. Triangle inequality postulate 7. If the measures of two angles of a triangle are unequal, then the lengths of the sides opposite these angles are unequal and the longer side lies opposite the larger angle. 8. A whole is greater than any of its parts. 9. Statements Reasons. AE BD, EC DC. Given. 2. AE + EC BD + DC 2. If unequal quantities are added to unequal quantities in the same order, the sums are 3. AC = AE + EC, 3. Partition BC = BD + DC 4. AC BC 4. Substitution 5. B A 5. If the lengths of two sides of a triangle are unequal, then the measures of the angles opposite these sides are unequal and the larger angle lies opposite the longer side. 0. Statements Reasons. AD DC. Given. 2. m ACD m CAD 2. If the lengths of two sides of a triangle are unequal, then the measures of the angles opposite these sides are unequal and the larger angle lies opposite the longer side. 3. ABC CDA 3. Given. 4. ACD BAC 4. Corresponding parts of congruent triangles are 5. m ACD m BAC 5. Definition of 6. m BAC m CAD 6. Substitution Postulate. 7. AC does not 7. Definition of angle bisect A. bisector.. Statements Reasons. ABC is isosceles. Given. with CA>CB. 2. m CAB m CBA 2. Base angles of an isosceles triangle are equal in measure. 3. m CBA m DBA 3. A whole is greater than any of its parts. 4. m CAB m DBA 4. Substitution 5. DB DA 5. If the measures of two angles of a triangle are unequal, then the lengths of the sides of opposite these angles are unequal and the longer side is opposite the larger angle. 2. Statements Reasons. RST is isosceles. Given. with RS ST. 2. m SRT m STR 2. Base angles of an isosceles triangle are equal in measure. 3. SRT and SRP 3. Definition of an form a linear pair. eterior angle of a STR and STQ triangle. form a linear pair. (Cont.) 282

Statements Reasons 4. m SRP m STQ 4. If two angles are congruent, then their supplements are 3. ADB 7 units; AEFGHIB 7 units. The length of one side of a triangle is less than the sum of the lengths of the other two sides. Therefore, CB AD and AC CB AC AD. We know that AC 2 and AD 4, so AC CB 7. Thus, ACB ADB. a. ACB is the shortest path. b. ADB and AEFGHB are the longest paths. Eploration (page 287). a. Proofs will vary. Let ABC and DEF be two triangles with AB>DE, BC>EF, and m B m E. Draw a point G on AC such that ABG E. Then ABG DEF by SAS. Corresponding parts of congruent triangles are congruent, so AG>DF and AG DF. Since a whole is greater than any of its parts, AC AG. By the substitution postulate, AC DF. b. Answers will vary. 2. a. Proofs will vary. Let ABC and DEF be two triangles with AB>DE, BC>EF, and AC DF. Draw a point G on AC such AG DF. Then ABG DEF by SSS. Corresponding parts of congruent triangles are congruent, so ABG E and m ABG m E. Since a whole is greater than any of its parts, m ABC m ABG. By the substitution postulate, m ABC m E. b. Answers will vary. Cumulative Review (pages 288 289) Part I. 3 2. 4 3. 3 4. 5. 4 6. 3 7. 8. 9. 4 0. 2 Part II. Yes. A statement and its contrapositive are logically equivalent. Therefore, If our meeting is not cancelled, then the snow does not continue to fall is true. By the law of detachment, since it is true that our meeting is not cancelled, we can conclude that snow does not continue to fall. 2. A ( 2, 4), B ( 2, 0), C (0, ) T 4, 5 (0, 3) ( 4, 2), r y= ( 4, 2) ( 2, 4) T 4, 5 (4, 3) (0, 2), r y= (0, 2) ( 2, 0) T 4, 5 (3, 5) (, 0), r y= (, 0) (0, ) Part III 3. Statements Reasons. AP BP. Assumption. 2. AP>BP 2. Definition of congruent segments. 3. PR bisects ARB. 3. Given. 4. R is the midpoint of 4. Definition of ARB. bisector. 5. AR>BR 5. Definition of midpoint. 6. PR>PR 6. Refleive property. 7. APR BPR 7. SSS. 8. ARP BRP 8. Corresponding parts of congruent triangles are 9. PR'ARB 9. If two lines intersect to form congruent adjacent angles, then they are perpendicular. 0. PR is not to ARB. 0. Given.. AP BP. Contradiction. 4. Statements Reasons. AC h bisectors DAB,. Given. CA h bisects DCB. 2. DAC BAC, 2. Definition of angle DCA BCA bisector. 3. AC>AC 3. Refleive property. 4. DAC BAC 4. ASA. 5. B D 5. Corresponding parts of congruent triangles are Part IV 5. m PTR m QTS because they are vertical angles, so y. m PTR m RTQ 80 since they are supplements, so 2 y 80 2 80 4 80 45 m PTR m QTS 45, m RTQ m PTS 2(45) 45 35 283

6. a. m BCA m DCB 80, so 6 8 4 2 80 m BCA 6(6) 8 04, m DCB 4(6) 2 76 0 20 80 0 60 6 b. m A m B m BCA; we are given that m A m B. Since m B m DCB and m DCB m BCA, by the transitive property of inequality, m B m BCA. c. BC AC AB; if the measures of two angles of a triangle are unequal, then the lengths of the sides opposite these angles are unequal and the longer side lies opposite the larger angle. 8- The Slope of a Line (page 295). Delta y means change of position with respect to the y-ais. 2. No. No slope is not the same as zero slope. Zero slope is a well-defined number. No slope occurs when the formula for the slope of a line, m5 y 2 2 y, is undefined. 3. b. m c. Slant upward 4. b. m 3 c. Slant upward 5. b. m 2 c. Slant downward 6. b. m c. Slant upward 7. b. m 2 2 c. Slant downward 8. b. No slope c. Vertical 9. b. m c. Slant downward 0. b. m 0 c. Horizontal. b. m 2 c. Slant upward 2. y O 2 2 (0, ) Chapter 8. Slopes and Equations of Lines 4. 5. 6. ( 4, 5) ( 3, 2) y O y y O O (2, 5) 3. y 7. y ( 4, 7) (, 3) O O 284