Triangle Congruence. Solutions Key. 59 Holt McDougal Geometry ARE YOU READY? CHECK IT OUT!

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1 CHAPTER Triangle Congruence Solutions Ke ARE YOU READY? 1. F. D 3. B. A 5. E Check students drawings. 1. 9_ + 7 = = 18 = _ (18) 9 = _ 3 = _ _ 3 3 = = _ 3 9_ _ 5 = 1_ 5 + 1_ = _ 13 5 = 3_ = 5-1_ = - 1 = 7_ = 3 1_ 17. Twice is 9 ft. = Price r is price p less 5. r = p Congruence and Transformations 16. t is 3 times m. t = 3m twice is = Half j is b plus 5 oz. 1_ j = b + 5 CHECK IT OUT! 1. D (3, 9), E (3, ), F (9, 0); dilation with scale factor D (1, 3) 0 - E (1, -) - -8 D (3, 9) F (3, 0) 6 E (3, ) F (9, 0) The triangles are congruent because ABC can be mapped to PQR b a rotation: (, ) (-, ). - Q(0, 3) R(-3, ) P(1, ) C(, 3) B(3, 0) A(, -1) 3. The triangles are cogruent because ABC can be mapped to A B C b a translation (, ) ( + 5, + ); and then A B C can be mapped to ABC b a reflection across the -ais Translation B(-, 1) A (1, 0) - A(-,-) P(1, 0) C(-,-) B (3, 3) C (3, 0) R(3, 0) Q(3, -3) Reflection. Possible answer: repeated horizontal reflections 59 Holt McDougal Geometr

2 THINK AND DISCUSS 1. output: ( + 5, - ); The image and preimage have the same size and shape. The image is 5 units right and units down from the preimage.. dilation with scale factor k 1 3. dilation with scale factor k < 1. Congruence Transformations Not Congruence Transformations 5. This transformation results in a 90 rotation clockwise with a center of rotation (0, 0). The -coordinates of the image are the -coordinates of the preimage. The -coordinates of the image are the additive inverses of the -coordinates of the preimage. The vertices of the new rectangle are: L (1, -3), M (, -3), N (, -5), O (1, -5). 6 M (3, ) N (5, ) translation: (, ) ( + a, + b) reflections: (, ) (-, ), (, ) (, -) rotations: (, ) (, -), (, ) (-, ), (, ) (-, -) dilation: (, ) (k, k), k > 0 L (3, 1) O (5, 1) EXERCISES GUIDED PRACTICE 1. are not. isometr, rigid transformation 3. The transformation results in a reflection across the -ais. The coordinates of the image are found b taking the -coordinate of each original point and the additive inverse of the -coordinate. The vertices of the new triangle are: A (, -1), B (5, -), C (5, -1). A(, 1) A (, -1) B (5, -) B(5, ) C(5, 1) 6 C (5, -1). This transformation results in a dilation with a scale factor of 3 and a center of (0, 0). The coordinates of the image are found b multipling the and coordinates of each original point b 3. The vertices of the new triangle are: P (, 3), Q (-3, 6), R (0, 3). Q (-3, 6) P (, 3) 6 P(-, 1) R(0, 1) R (0, 3) Q(-1, ) L (1, -3) - O (1, -5) M (, -3) N (, -5) 6. This transformation results in a shift 3 units to the left and units up. The -coordinates of the image are found b subtracting 3 from the -coordinates of each original point. The -coordinates of the image are found b adding to the -coordinates of each original point. The vertices of the new triangle are: D (1, 1), E (, 5), F (, 1) D (1, 1) - 0 E (, 5) F (, 1) E(7, 3) 8 D(, -1) F(7, -1) 7. The rectangles are not congruent. WXYZ is a result of dilating ABCD with a scale factor of 0.5. If the scale factor, k, does not equal 1, the two figures are not congruent. (, ) (0.5, 0.5). -8 B(-, 6) C(, 6) A(-, ) X(-, 3) D(, ) Y(1, 3) W(-, ) Z(1, ) Holt McDougal Geometr

3 8. The triangles are congruent. Triangle TUV is a result of rotating triangle ABC around the origin. The mapping does not require a dilation. The scale factor is 1, so the figures are congruent. (, ) (-, -). V(1, 1) C(-1, -1) 6 B(-, -1) A(-, -) - T(, ) U(, 1) 9. The triangles are congruent since triangle MNO is a result of shifting JKL two units down and reflecting the triangle across the -ais. The mapping does not require a dilation. The scale factor is 1. (, ) (-, - ). J(-5, ) L(-, ) K(-, 5) N(, 3) O(, 0) M(5, 0) 10. The triangles are congruent because triangle DEF can be mapped to triangle D E F b a rotation: (, ) (-, -), and then triangle D E F can be mapped to triangle XYZ b a translation: (, ) ( +, - 1). - - F(-1, -) E(-, -) D(-1, -5) - Z(3, 1) X(3, ) 6 6 Y(6, 3) 11. Repeated horizontal reflections and horizontal translations create the wallpaper pattern. The large flower at the top is translated right and left, while the stem, leaves, smaller flowers, and background design are reflected to the left and right to create an image that is congruent to the preimage. 1. Answers will var. PRACTICE AND PROBLEM SOLVING 13. This transformation results in a shift 5 units to the right and units down. The -coordinates of the image are found b adding 5 to the -coordinates of each original point. The -coordinates of the image are found b subtracting from the -coordinates of each original point. The vertices of the new triangle are: G (9, -5), H (1, -1), I (1, -5) H(7, 3) G(, -1) I(7, -1) H (1, -1) G (9, -5) I (1, -5) 1. The transformation results in a reflection across the -ais. The coordinates of the image are found b taking the -coordinate of each original point and the additive inverse of the -coordinate. The vertices of the new triangle are: P (-3, ), Q (, ), R (-3, 5). R (-3, 5) 6 R(3, 5) Q (, ) P (-3, ) P(3, ) Q(6, ) This transformation results in a dilation with a scale factor of 1.5 and a center of (0, 0). The coordinates of the image are found b multipling the and coordinates of each verte in the preimage b 1.5. The vertices of the new triangle are: L (-1.5, 6), M (, 6), N (,.5). M (, 6) N (,.5) M(-, ) N(-, 3) L (-1.5, 6) L(-1, ) 61 Holt McDougal Geometr

4 16. This transformation results in a 90 rotation counterclockwise with a center of rotation (0, 0). The -coordinates of the image are the additive inverses of the -coordinates of the preimage. The -coordinates of the image are the -coordinates of the preimage. The vertices of the new rectangle are: A (, -7), B (-, -7), C (, -), D (-, -). A(-7, 6) C(-, 6) B(-7, ) D(-, ) C (, -) D (-, -) - A (, -7) B (-, -7) 17. This transformation results in a shift 1 unit to the left and 1 unit up. The -coordinates of the image are found b subtracting 1 from the -coordinates of each original point. The -coordinates of the image are found b adding 1 to the -coordinates of each original point. The vertices of the new triangle are: N (0, -1), O (-1, 5), P (1, 5) O (-1, 5) P (1, 5) O(0, ) P(, ) N (0, -1) N(1, -) 18. This transformation results in a 180 rotation with a center of rotation (0, 0). The -coordinates of the image are the additive inverses of -coordinates of the preimage. The -coordinates of the image are the additive inverses of the -coordinates of the preimage. The vertices of the new triangle are: W (-5, -), X (-, -), Y (-5, -5). W(5, ) X(, ) W (-5, -) X (-, -) Y (-5, -5) - Y(5, 5) 19. The rectangles are not congruent because rectangle JKLM can be mapped to rectangle ABCD b a dilation with scale factor k 1: (, ) (0.5, 0.5). -8 K(-, 6) L(, 6) J(-, ) M(, ) B(-, 3) C(1, 3) A(-, ) D(1, ) The triangles are congruent. Triangle XYZ is a result of rotating triangle PQR 180 around the origin using the transformation (, ) (-, -). There is no dilation needed for the mapping. The scale factor is 1. Z(1, 1) R(-1, -1) Q(-, -1) - P(-, -) - X(, ) Y(, 1) 1. Yes, the triangles are congruent because triangle EFG can be mapped to triangle UVW b a translation: (, ) (, + 3). W(-3, 6) G(-3, 3) U(-1, ) 6 E(-1, -1) - - V(, 5) F(, ) 6 6. The triangles are congruent because triangle DFG can be mapped onto triangle XYZ through two transformations, neither requiring a dilation. The scale factor will be one. First the triangle is shifted units to the right: (, ) ( +, ). This triangle is then reflected across the -ais: (, ) (, -). E(-, 1) X(-1, 1) D(-5, -1) - Y(, -1) F(, -1) Z(6, 1) 8 6 Holt McDougal Geometr

5 3. The triangles are congruent because triangle ABC can be mapped onto triangle DEF through two transformations, neither requiring a dilation. The scale factor will be one. First the triangle is rotated 90 clockwise around (0, 0): (, ) (, -). This triangle is then shifted two units to the left: (, ) ( -, ). C(6, 0) A(, -1) D(-3, -) - B(, -) E(-, -) - F(-, ). The triangles are congruent because triangle PQR can be mapped onto triangle GHR through two transformations, neither requiring a dilation. The scale factor will be one. First the triangle is shifted 3 units to the right: (, ) ( + 3, ). This triangle is then rotated 180 around the origin: (, ) (-, -). Q(-8, 7) -10 P(-7, 3) -8 - R(-, 7) I(1, -7) G(, -3) H(5, -7) 5a. The pattern was created with a 90 counterclockwise rotation and translation to the right and down. Then another 90 rotation and a translation right and down. This then repeats to fill the rectangular quilt area. b. The thin rectangles are congruent, as are the quilt block squares. No dilation was needed to do an transformation. The scale factor remains 1 throughout the quilt. c. The quilt would look more like a checkerboard pattern. 6. The transformation results in a shift 3 units to the left and units up. The -coordinates of the image are found b taking the -coordinate of each original point and subtracting 3. The -coordinates of the image are found b taking the -coordinate of each original point and adding. The vertices of the new triangle are: A (-1, 0), B (3, 0), C (-1, -) - A (-1, 0) C (-1, -) B (3, 0) A(, -) - B(6, -) C(, ) 7. This transformation results in a 90 rotation clockwise with a center of rotation (0, 0). The -coordinates of the image are the -coordinates of the preimage. The -coordinates of the image are the additive inverses of the -coordinates of the preimage. The vertices of the new rectangle are: X (, 6), Y (, -), Z (-, 0) X(, ) Z (-, 0) Z(0, -) - X (, 6) Y(, ) 6 Y (, -) 8. Eli rotated the design 90 around the lower left hand corner. He then translated the image to the right and down. There was no dilation. The two black shapes are congruent. 9. The transformation results in a reflection across the -ais and a shift 3 units to the right. The -coordinates of the image are found b taking the -coordinate of each original point and adding 3. The -coordinates of the image are found b taking the additive inverses of the -coordinate of each original point. The vertices of the new triangle are: A (, -5), B (, -), C (-1, ) A(-1, 5) B(1, ) C (-1, ) C(-, -) - B (, -) - A (, -5) 63 Holt McDougal Geometr

6 30. This transformation results in a dilation of scale factor 3 and center (0,0) and then 90 rotation clockwise with a center of rotation (0, 0). The -coordinates of the image are -3 times the -coordinates of the preimage. The -coordinates of the image are 3 times the -coordinates of the preimage. The vertices of the new rectangle are: X (, 3), Y (-3, 1), Z (3, 9) Y (-3, 1) X (, 3) X(1, ) Z (3, 9) Y(, 1) Z(3, -1) 31. Starting at the upper left, Frank reflected his shape horizontall. Then he rotated 90 counterclockwise and translated it to the right and up. Then he reflected it verticall. This whole row was then rotated 180 and translated down. 3. Seth will reflect the triangle horizontall over the top point of the roal blue diamond. Then he will reflect the top corners verticall over the side points of the diamond. 33. Dave is correct. Erin states that the triangle was rotated, before the translation, but the triangle was actuall reflected across the -ais before the translation was applied. 3. When ou reflect a polgon, ou use (, ) (-, ) to reflect across the -ais and (, ) (, -) to reflect across the -ais. When ou rotate a polgon, ou use (, ) (, -) to rotate 90 clockwise about (0, 0), (, ) (-, ) to rotate 90 counterclockwise about (0, 0) and (, ) (-, -) to rotate 180 about (0, 0). For the 90 rotations, the and values are interchanged, which does not occur when reflecting a polgon. 35. The lengths of the sides of a figure dilated with a scale factor of will be times as large as the length of the sides of the same figure dilated with a scale factor of 0.5. A scale factor of increases the figure to twice its original size. A scale factor of 0.5 decreases the figure to half its original size. The area of the larger dilation will be 16 times larger than the area of the smaller dilation. STANDARDIZED TEST PREP 36. B; Dilate triangle DEF b a factor of. D(-5, -) (-10, -) E(-, -) (-, -) F(-, -5) (-8, -10) Then reflect across the -ais. (-10, -) X(10, -) (-, -) Y(, -) (-8, -10) Z(8, -10) 37. D; The transformation keeps the -coordinate the same and multiplies the -coordinate b -1. A(1, 1) A (1, -1) B(6, ) B (6, -) C(8, -) C (8, ) CHALLENGE AND EXTEND 38a. Yes, the squared S-shapes are congruent, as are the jagged lines, which are each made of 10 pieces of stone. The S-shapes and jagged lines in one row are reflected verticall and then the entire row is translated up and over so the designs fit the space more tightl. b. The S-shapes would be on their side and look more like N-shapes. The would not connect with the jagged lines as well. Instead of the two rows fitting together like interlocking pieces, the rows would be eactl the same and create repeating diagonal lines in the same direction. - - Classifing triangles CHECK IT OUT! 1. FHG and EHF are complementar. m FHG + m EHF = 90 m FHG + 30 = 90 m FHG = 60 All are equal. So FHG is equiangular b definition.. AC = AB = 15 No sides are congruent. So ACD is scalene. 3. Step 1 Find the value of. FG GH FG = GH 3 - = = + 7 = 7 Step Substitute 7 for. FG = 3 - GH = + 3 = 3(7) - = 17 = (7) + 3 = 17 FH = 5-18 = 5(7) - 18 = 17 a. P = 3(7) = 1 in = 16_ 1 triangles b. P = 3(10) = 30 in = 3 1_ 3 3 triangles Think and discuss 1. DE, EF, E; EF, FD, F; FD, DE, D 6 Holt McDougal Geometr

7 . Possible answer: 3. No; all 3 in an acute must be acute, but the do not have to have the same measure; possible answer:. In an equil. rt., all 3 sides have the same length. B the Pth. Thm., the 3 side lengths are related b the formula c = a + b, making the hp. c greater than either a or b. So the 3 sides cannot have the same length. 5. Classification B sides: equil.: 3 sides isosc.: at least sides scalene: no sides Eercises guided practice B : acute: 3 acute equiangular: 3 acute rt.: 1 rt. obtuse: 1 obtuse 1. An equilateral triangle has three congruent sides.. One angle is obtuse and the other two angles are acute. 3. DBC is a rt.. So DBC is a rt... ABD and DBC are supp. ABD + DBC = 180 ABD + 90 = 180 ABD = 90 ABD is a rt.. So ABD is a rt.. 5. m ADC = m ADB + m BDC = = 101 ADC is obtuse. So ADC is an obtuse. 6. EG = = 6, EH = 8, GH = 8 EH GH Eactl two sides are, so EGH is isosc. 7. EF = 3, EH = 8, FH = 7. No sides are congruent, so EFH is scalene. 8. GF = 3, GH = 8, FH = 7. No sides are congruent, so HFG is scalene. 9. Step 1 Find. 6 = + 1 = 1 = 6 Step Find side lengths. is equilateral, so all three side lengths = 6 = Step 1 Find = +. = = 0.7 Step Find side lengths. +. = = = (0.7) = = = (0.7) = = Perimeter is P = = 7.5 cm = 6 _ 3 earrings The jeweler can make 6 earrings. practice and problem solving 1. m BEA = 90 ; rt. 13. m BCD = = 10 ; obtuse 1. m ABC = = 60 m ABC = m ACB = m BAC; equiangular 15. PS ST PT ; equilateral 16. PS RS, so PS = RS = 10; RP = 17; isosc. 17. RT = = 0, RP = 17, PT = 10; scalene 18. Step 1 Find z. 3z - 1 = z + 5 3z = z + 6 z = 6 z = 3 Step Find side lengths. z + 5 = = 8 3z - 1 = 3(3) - 1 = 8 z - = (3) - = 8 0a. Check students drawings. XY, YZ, XZ, X, Y, Z 19. Step 1 Find = = = 5. = 0.9 Step Find side lengths = 8(0.9) + 1. = = = (0.9) = = 8.6 b. Possible answer: scalene obtuse 1. PQ + PR + QR = 60 PQ + PQ + _ PQ = _ PQ = 60 3 PQ = 3_ (60) = 18 ft 10 PR = PQ = 18 ft QR = _ PQ = _ (18) = ft = 1_ ; complete trusses 3.. Not possible: an equiangular has onl acute Not possible: an equiangular must also be equilateral Holt McDougal Geometr

8 9. Let represent each side length. + + = = 105 = 35 in. 30. AB AC, so is isosc. BAC and CAD are supp., and CAD is acute; so BAC is obtuse. ABC is isosc. obtuse. 31. AC CD and m ACD = 90. ACD is isosc. rt. 3. ( - 1) + ( - 1) + = = 3 9 = 36 = 33a. E nd Street side = 1_ (Broadwa side) - 8 = 1_ (190) - 8 = 87 ft 5th Avenue side = (E nd Street side) - 1 = (87) - 1 = 173 ft b. All sides are different, so is scalene. 3. No; es; not ever isosc. is equil. because onl of the 3 sides must be. Ever equil. has 3 sides, and the def. of an isosc. requires that at least sides be. 35. S; equil, acute 36. S; scalene, acute 37. A; 3 congruent sides, so alwas satisfies isosceles classification 38. s = P_. The perimeter of an equil. is 3 times the 3 length of an 1 side, or P = 3s. Solve this formula for s b dividing both sides b Check students constructions. 0a. DE = AD + AE = 5 + ( _ 10 ) = = 50 DE = ÇÇ 50 = 5 Ç cm Think: CE DE. CE = DE = 5 Ç cm b. Think: DE bisects AEF. m DEF = 1_ m AEF = 1_ (90) = 5 Think: CEF DEF, so m CEF = 5. m DEC = m DEF + m CEF = = 90 c. CE = DE and m DEC = 90 isosc. ; rt. test prep 1. D 3s = P 3s = 36 _ 3 s = 1_ 3( 36 + _ 3 ) = 1 _ 9 in.. F B graphing, RT RS ST, so RST is isosc.. 3 P = AB + BC + AC = 1_ + 1_ + 5_ - + 1_ + 1_ = ( 1_ _ ) + 1_ + 5_ + 1_ = 3 challenge and etend 3. D LMN has no rt.. 5. It is an isosc. since sides of the have length a. It is also a rt. since sides of the lie on the coord. aes and form a rt.. 6. Statements Reasons 1. ABC is equiangular.. A B C. Def. of equiangular 3. EF ǁ AC 3. Given. BEF A,. Corr. Post. BFE C 5. BEF B, 5. Trans. Prop. of BFE B 6. BEF BFE 6. to the same are. 7. EFB is equiangular. 7. Def. of equiangular 7. Think: Each side has the same measure. Use the epression + 10 for this measure. 3( + 10) = = 1 3 = -9 = Step 1 Find. Think: Average of + 1, 3 +, and 8-16 is. 1_ ( ) = 3 1_ (1) = 3 = = 6 Step Find side lengths. + 1 = = = 3(6) + = 8-16 = 8(6) - 16 = 3 longest side - average = 3 - = 8 66 Holt McDougal Geometr

9 -3-3 Angle relationships in triangles CHECK IT OUT! 1. Step 1 Find m NKM. m KMN + m MNK + m NKM = m NKM = m NKM = 180 m NKM = Step Find m MJK. m JMK + m JKM + m MJK = m MJK = m MJK = 180 m MJK = 3 a. Let acute be A, B, with m A = m A + m B = m B = 90 m B = 6.3 b. Let acute be C, D, with m C =. m C + m D = 90 + m D = 90 m D = (90 - ) c. Let acute be E, F, with m E = 8 5. m E + m F = m F = 90 5 m F = 1 3_ 5 3. m ACD = m ABC + m BAC 6z - 9 = 90 + z + 1 z = 100 z = 5 m ACD = 6z 9 = 6(5) 9 = 11. P T m P = m T = = -3 = 16 So m P = = 3. Since m T = m P, m T = 3. THINK AND DISCUSS Since 3 and are supp., m 3 + m = 180 b def = 180 b the Sum Thm. B the trans. Prop. of =, m 3 + m = m 1 + m + m 3. Subtract m 3 from both sides. Then m = m 1 + m.. ; 6 3. Theorem Words Diagram Sum Thm. Et. Thm. Third Thm. Eercises guided practice The sum of the measures of the int. of a is 180. The measure of an et. of a is = to the sum of the measures of its remote int.. If of 1 are to of another, then the third pair of are. m 1 + m + m 3 = m = m 1 + m Possible answers: think out of the wa. Eterior is net to E. So the remote interior are D and F. 3. auiliar lines. Think: Use Sum Thm. 180 = = = 10 = Deneb: = 3(17) + 13 = 6 Altair: + = (17) + = 36 Vega: 5-5 = 5(17) - 5 = m = 90 m = _ + m = 90 3 m = 65 1_ 3 9. m M + m N = m NPQ = = 8 5 = 5 = 9 m M = = 3(9) + 1 = m K + m L = m HJL = = 91 = 7 m L = 6-1 = 6(7) - 1 = m A + m B = m B = 117 m B = 5 m A + m B + m BCA = m BCA = 180 m BCA = m = 90 m = (90 - ) 67 Holt McDougal Geometr

10 1. C F m C = m F = = 5 m C = = 100 m F = m C = C Z m C = m Z + 7 = 3( + 5) + 7 = = 8 m C = + 7 = (8) + 7 = 39 m Z = m C = 39 practice and problem solving 15. m A + m B + m P = m P = m P = 180 m P = m = 90 m = 13 3_ m = 90 m = Think: Use Et. Thm. m W + m X = m XYZ = = = = 1 m XYZ = = 15(1) - 18 = S U m S = m U 5-11 = + 9 = 0 m S = 5-11 = 5(0) - 11 = 89 m U = m S = m = 90 m = (90 - ) 0. Think: Use Et. Thm and subst. m C = m D. m C + m D = m ABD m D = m ABD (6-5) = = = 11 m C = m D = 6-5 = 6(11) - 5 = Think: Use Third Thm. N P m N = m P 3 = = -1 = 16 m N = 3 = 3(16) = 8 m P = m N = 8. Think: Use Third Thm. Q S m Q = m S = = m Q = = (6) = 18 m S = m Q = Think: Use Sum Thm. m 1 + m + m 3 = = = 180 = 15 m 1 = = 15 m = = 60 m 3 = 7 = 105. Statements Reasons 1. DEF with rt. F. m F = 90. Def. of rt. 3. m D + m E + m F = Sum Thm.. m D + m E Subst. = m D + m E = Subtr. Prop. 6. D and E are comp. 6. Def. of comp. 5. Proof 1: Statements Reasons 1. ABC is equiangular. m A = m B = m C. Def. of equilangular 3. m A + m B + m C 3. Sum Thm. = 180. m A + m A + m A = 180 m B + m B + m B = 180 m C + m C + m C = m A = 180, 3m B = 180, 3m C = m A = 60, m B = 60, m C = 60. Subst. prop 5. Simplif. 6. Div. Prop. of = Proof : A, B, and C are all congruent, so their measures are equal. The sum of the three measures is 180, b Sum Thm. Therefore, 3 (common measure) = 180. So the common measure is 60. That is, m A = m B = m C = Step 1 Write an equation. m 1 = 1 1_ m Step Since the acute of a rt. are comp. write and solve another equation. m 1 + m = _ m + m = 90 9_ m = 90 m = _ (90) = 0 9 Step 3 Find the larger acute, m 1. m 1 = 1 1_ m = 5_ (0) = Holt McDougal Geometr

11 7. A B C D E Statements 1. ABC, DEF, A D, B E F Reasons. m A + m B + m C = 180. Sum Thm. 3. m C = m A - m B 3. Subtr. Prop. of =. m D + m E + m F = 180. Sum Thm. 5. m F = m D - m E 5. Subtr. Prop. of = 6. m A = m D, m B = m E 6. Def. of 7. m F = m A - m B 7. Subst. 8. m F = m C 8. Trans. Prop. of = 9. F C 9. Def. of 8. Statements Reasons 1. ABC with et. ACD. m A + m B + m ACB = 180. Sum Thm. 3. m ACB + m ACD = Lin. Pair Thm.. m ACD = m ACB. Subtr. Prop. of = 5. m ACD = (m A + m B + m ACB) - m ACB 5. Subst. 6. m ACD = m A + m B 6. Simplif. 9. Think: Use Alt. Int. Thm. m WUX + m UXZ = 180 m WUX + 90 = 180 m WUX = 90 So UWX is a rt.. m UXW + m XWU = 90 m UXW + 5 = 90 m UXW = XWU, UWY, and YWV are supp.. m XWU + m UWY + m YWV = m UWY + 78 = 180 m UWY + 13 = 180 m UWY = Think: Use Third Thm. WUY ZXY UYW XYZ WZX UWY m WZX = m UWY = 8 3. XYZ and WZX are acute in a rt.. m XYZ + m WZX = 90 m XYZ + 8 = 90 m XYZ = 33. Let 1,, and 3 be internal. Let, 5, and 6 be eternal. Think: Use Et. Thm. m = m 1 + m m 1 = m = 60 So m = = 10. Likewise, m 5 = m 6 = 10. Et. sum = m + m 5 + m 6 = Think: Use Third Thm. SRQ RST m SRQ = m RST = Let acute measures be and. + = 90 5 = 90 = 18 Smallest measure is = a. hpotenuse b = 180 c. + = 90 and are comp. measures. d. z = e. + = = 90 = 53 z = + 90 z = z = 17 The et. at the same verte of a are vert.. Since vert. are, the et. have the same measure. 38. Statements Reasons 1. AB BD, BD CD, A C. ABD and CDB are rt.. Def. of lines 3. m ABD = m CBD 3. Def. of rt.. ABD CDB. Rt. Thm. 5. ADB CBD 5. Third Thm. 6. AD ǁ CB 6. Conv. of Alt. Int. Thm. 39. Check students sketches. Et. measures = sums of remote int. measures: 155, 65, and 10. 0a. m FCE = 1 m DCE = 1 (90) = 5 m FCB = 1 m FCE = 1 (5) =.5 b. m CBE + m BEC + m BCE = 180 m CBE = 180 m CBE = 180 m CBE = Holt McDougal Geometr

12 test prep 1. C 18 = 71 + = 57. F (s + 10) = 180 s + 13 = 180 s = 6 s = 3 3. D m A + m B = m BCD m B = m BCD - m A. Let, 3, and represent the measures. The sum of the measures of a is 180, so = 180. Solving the eqn. for the value of, ields = 0. Find each measure b subsituting 0 for in each epression. = (0) = 0; 3 = 3(0) = 60; = (0) = 80. Since all of the measure less than 90, the are all acute b def. Thus the is acute. challenge and etend = ( + 7) + (61 - ) 117 = = = 7 or A rt. is formed. The same-side int. are supp., so the formed b their bisectors must be comp. That means the remaining of the must measure Since an et. is = to a sum of remote int., it must be greater than either. Therefore, it cannot be to a remote int.. 8. Possible sets of measures: (30, 30, 10), (30, 60, 90), (60, 60, 60) Probabilit = _ 3 9. Let m A =. m B = 1 1_ ( ) - 5 m C = 1_ ( ) - 5 m A + m B + m C = _ ( ) _ ( ) - 5 = = = 190 = 38 m A = = Congruent Triangles CHECK IT OUT! 1. Angles: L E, M F, N G, P H Sides: LM EF, MN FG, NP GH, LP EH a. AB DE - = 6 = 8 = b. Since the acute of a rt. are comp. m B + m C = m C = 90 m C = 37 F C m F = m C = Statements Reasons. 1. A D. BCA ECD. Vert. are. 3. ABC DEC 3. Third Thm.. AB DE. Given 5. AD bisects BE, and BE bisects AD. 5. Given 6. BC EC, AC BC 6. Def. of bisector 7. ABC DEC 7. Def. of Statements 1. JK ǁ ML. KJN MLN, JKN LMN Reasons. Alt. Int. Thm. 3. JNK LNM 3. Vert. Thm.. JK ML. Given 5. MK bisects JL, and JL bisects MK. 5. Given 6. JN LN, MN KN 6. Def. of bisector 7. JKN MLN 7. Def. of THINK AND DISCUSS 1. Measure all the sides and all the. The trusses are the same size if all the corr. sides and are.. PQR LMN Angles: P L Q M R N Sides: PQ LM QR MN PR LN Eercises guided pratice 1. You find the and. B sides that are in the same, or matching, places in the. 3. LM. RT 5. M 6. NM 7. R 8. T 9. JK FG 10. G K JK = FG 3-15 = 1 3 = 7 = 9 m G = m K - 0 = 108 = 18 = 3 KL = = 9 70 Holt McDougal Geometr

13 11. Statements Reasons 1. AB ǁ CD.. ABE CDE, BAE DCE 3. AB CD. E is the mdpt. of AC and BD 5. AE CE, BE DE. Alt. Int. Thm. 3. Given. Given 5. Def. of mdpt. 6. AEB CED 6. Vert. Thm 7. ABE CDE 7. Def. of practice and problem solving 1. Statements Reasons 1. UST RST, U R. STU STR. Third Thm. 3. SU SR 3. Given. ST ST 5. TU TR. Refle. Prop. of 5. Given 6. RTS UTS 6. Def. of 13. LM 1. CF 15. N 16. D 17. ADB CDB m ADB = m CDB + 10 = 90 = 80 = AB CB AB = CB - 7 = 1 = 19 m C = + 11 = Statements Reasons 1. N R. MP bisects NMR. Given 3. NMP RMP 3. Def. of bisector. NPM RPM. Third Thm. 5. P is the mdpt. of NR 5. Given 6. PN PR 7. MN MR 8. MP MP 6. Def. of mdpt. 7. Given 8. Refle. Prop. of 9. MNP MRP 9. Def. of 1. GSR KPH, SRG PHK RSG HPK, 3. AB DE AB = DE - 10 = + 0 = 30 AB = - 10 = (30) - 10 = BC QR BC = QR = = BC = = 6() + 5 = 17 6a. KL ML b the def. of a square.. RVUTS VWXZY. L P m L = m P + 10 = = m L = + 10 = = 19 b. Statements Reasons 1. JKLM is a square.. KL ML. Def. of a square 3. JL and MK are 3. Given bisectors of each other.. MN KN. Def. of bisector 5. NL NL 5. Refle. Prop. of 6. MNL and KNL are 6. Def. of rt.. 7. MNL KNL 7. Rt. Thm. 8. NML NKL 8. Given 9. NLM NLK 9. Third Thm. 10. NML NKL 10. Def. of 0. Statements Reasons 1. ADC and BCD are rt.. ADC BCD. Rt. Thm. 3. DAC CBD 3. Given. ACD BDC. Third Thm. 5. AC BD, AD BC 5. Given 6. DC DC 6. Refle. Prop. of 7. ADC BCD 7. Def. of 71 Holt McDougal Geometr

14 7. A B D C Statements Reasons 1. BD AC. ADB and CDB are. Def. of rt.. 3. ADB CDB 3. Rt. Thm.. BD bisects ABC.. Given 5. ABD CBD 5. Def. of bisector 6. A C 6. Third Thm. 7. AB CB 7. Given 8. BD BD 8. Refle. Prop. of 9. D is the mdpt. of AC. 9. Given 10. AD CD 10. Def. of mdpt. 11. ABD CBD 1. Def. of 8. Possible answer: challenge and etend 35. P = TU + UV + VW + TW 19 = = = 30 = 5.5 Yes; UV = WV = 1.5, and UT = WT = 33. TV = TV b the Refle. Prop. of =. It is given that VWT VUT and WTV UTV. WVT = UVT b the Third Thm. Thus TUV TWV b the def. of. 36. E A m E = m A - 10 = 90 = 100 m D = m H = - 13 = (100) - 13 = Statements Reasons 1. RS RT ; S T. ST TS. Refle. Prop. of 3. T S 3. Sm. Prop. of. R R. Refle. Prop. of 5. RST RTS 5. Def. of READY TO GO ON? Section a Quiz.5 cm cm 3. cm cm 9. Solution A is incorrect. E M, so m E = Yes; b the Third Thm., K W, so all 6 pairs of corr. parts are. Therefore, the are. test prep 31. B Matching up, ABC FDE. 3. G N S m N = m S 6 = = = D m Y = (m X + m Z) = (m A + m C) = = J P = MN + NR + RM = SP + QP + SR + RQ = = 97 M R m M = m R 58 = 3-60 = 3 = 0 1. M : (5, ) (5 -, + 3); A (3, 5) M : (-3, ) (-3 -, + 3); B ( 5, 7) M : (-1, ) (-1 -, + 3); C ( 3, 3). M : (5, ) (5 -()); A (5, -) M : (-3, ) (-3 - ()); B (-3, -) M : (-1, ) (-1 - ()); C ( 1, 6) 3. M : (5, ) (-(), 5); A (-, 5) M : (-3, ) (-(), -3); B (-, -3) M : (-1, ) (-(), -1); C (6, -1). M : (5, ) (3(5), 3()); A (15, 6) M : (-3, ) (3(-3), 3()); B (-9, 1) M : (-1, ) (3(-1), 3()); C (-3, -18) 5. rt., since ACB is rt. 6. equiangular, since m BAD = = 60 = m B = m ADB 7. obtuse, since m ADE = m B + m BAD = isosc., since PQ = QR = 5, PR = equilateral, since PR = RS = PS = scalene, since PQ = 8.7, QS = = 10, PS = m M + m N = m NLK = = 6 = 8 m M = = 51 7 Holt McDougal Geometr

15 1. m C + m D = m ABC = = 15 = 7 m ABC = 0 15 = EF 1. JL 15. E 16. L 17. Statements Reasons AB È ǁ CD È. BAD CDA. Alt. Int. Thm. 3. AC CD, DB AB 3. Given.. ACD and DBA are rt.. Def. of 5. ACD DBA 5. Rt. Thm. 6. CAD BDA 6. Third Thm. 7. AB CD, AC DB 7. Given 8. AD DA 8. Refle. Prop. of 9. ACD DBA 9. Def. of TRiangle congruence: SSS and SAS CHECK IT OUT! 1. It is given that AB CD and BC DA. B the Refle. Prop. of, AC CA. So ABC CDA b SSS.. It is given that AB BD and ABC DBC. B Refle. Prop. of, BC BC. So ABC DBC b SAS. 3. DA = 3t + 1 = 3() + 1 = 13 DC = t - 3 = () - 3 = 13 m ADB = 3 m CDB = t = ( ) = 3 DA DC, DB DB, and ADB CDB So ADB CDB b SAS.. Statements Reasons 1. QR QS. ÈÈ QP bisects RQS. Given 3. RQP SQP 3. Def. of bisector. QP QP. Refle. Prop. of 5. RQP SQP 5. SAS Steps 1, 3, Think and Discuss 1. Show that all si pairs of corr. parts are ; SSS; SAS. The SSS and SAS Post. are methods for proving without having to prove of all 6 corr. parts. 3. SSS SAS How are the alike? Both posts. use sides and an included corr. part. Eercises guided practice 1. T How are the different? For SSS the included part is a side. For SAS the included part is an.. It is given that DA BC and AB CD. BD DB b the Refle. Prop. of. Thus ABD CDB b SSS. 3. It is given that MN MQ and NP QP. MP MP b the Rele. Prop. of. Thus MNP MQP b SSS.. It is given that JG LG, and GK GH. JGK LGH b the Vert. Thm. So JGK LGH b SAS. 5. When =, HI = GH = 3, and IJ = GJ = 5. HJ HJ b the Refle. Prop. of. Therefore, GHJ IHJ b SSS. 6. When = 18, RS = UT = 61, and m SRT = m UTR = 36. RT TR b the Refle. Prop. of. So RST TUR b SAS. 7. Statements Reasons 1. JK ML. JKL MLK. Given 3. KL LK 3. Refle. Prop. of. JKL MLK. SAS Steps 1,, 3 practice and problem solving 8. It is given that BC = ED = in. and BD = EC = 3 in. So b the def. of, BC ED, and BD EC. DC CD b the Refle. Prop. of. Thus BCD EDC b SSS. 9. It is given that KJ LJ and GK GL. GJ GJ b the Refle. Prop. of. So GJK GJL b SSS. 10. It is given that C and B are rt. and EC DB. C B b the Rt. Thm. CB BC b the Refle. Prop. of. So ECB DBC b SAS. 11. When = 3, NQ = NM = 3, and QP = MP =. So b the def. of, NQ NM, and QP MP. m M = m Q = 90, so M Q b the def. of. Thus MNP QNP b SAS. 1. When t = 5, YZ =, ST = 0, and SU =. So b the def. of, XY ST, YZ TU, and XZ SU. This XYZ STU b SSS. 73 Holt McDougal Geometr

16 13. Statements Reasons 1. B is mdpt. of DC. DB CB 3. AB DC. ABD and ABC are rt.. Def. of mdpt. 3. Given. Def. of 5. ABD ABC 5. Rt. Thm. 6. AB AB 6. Refle. Prop. of 7. ABD ABC 7. SAS Steps, 5, 6 1. SAS (with Refle. Prop of ) 15. SAS (with Vert. Thm.) 16. neither 17. neither 18a. To use SSS, ou need to know that AB DE and CB CE. b. To use SAS, ou need to know that CB CE. 19. QS = ÇÇÇÇ 1 + = Ç 5 SR = ÇÇÇÇ + 0 = QR = ÇÇÇÇ 3 + = ÇÇ 13 TV = ÇÇÇÇ 1 + = Ç 5 VU = ÇÇÇÇ + 0 = TU = ÇÇÇÇ 3 + = 13 ÇÇ The are b SSS. 0. AB = ÇÇÇÇ 1 + = ÇÇ 17 BC = ÇÇÇÇ + 3 = 5 AC = ÇÇÇÇ = ÇÇ 6 DE = ÇÇÇÇ 1 + = ÇÇ 17 EF = ÇÇÇÇ + 3 = 5 DF = ÇÇÇÇ + 0 = The are not. 1. Statements Reasons 1. ZVY WYV, ZVW WYZ. m ZVY = m WYV, m ZVW = m WYZ 3. m ZVY + m ZVW = m WYV + m WYZ. Def. of 3. Add. Prop. of =. m WVY = m ZYV. Add. Post. 5. WVY ZYV 5. Def. of 6. WV YZ 6. Given 7. VY YV 7. Refle. Prop. of 8. ZVY WYV 8. SAS Steps 6, 5, 7 3. b. 3.5 ft; b the Pth. Thm., BC 3.5 ft. Since the are congruent, EF BC AB = AC = = = 5.5 BC = DC + = = = 5.5 B the def. of, AB BD, and BC DC. AC AC b the Refle. Prop. of. Thus ABC ADC b SSS. 5. Measure the lengths of the logs. If the lengths of the logs in 1 wing deflector match the lengths of the logs in the other wing deflector, the will be b SAS or SSS. 6. Yes; if the have the same side lengths and the same included measure, the are b SAS. 7. Check students constructions; es; if each side is to the corr. side of the second, the can be in an order. test prep 8. C In I and III, two sides are congruent with an congruent angle in between so I and III are similar b SAS. 9. G SAS proves ABC ADC, so AB + BC + CD + DA = AB + CD + CD + AB = = 39.8 cm 30. A F and J are the included, so F J proves SAS. 31. J EF EH EF = EH + 7 = 6-11 = = 5.5 a. Measure AB and AC on 1 truss and measure DE and DF on the other. If AB DE and AC DF, then the trusses are b SAS. 7 Holt McDougal Geometr

17 challenge and etend 3. Statements Reasons 1. Draw DB. 1. Through an pts. there is eactl one line.. ADC and BCD are supp. 3. AD ǁ CB. Given 3. Conv. of Same-Side Int. Thm.. ADB CBD. Alt. Int. Thm. 5. AD CB 5. Given 6. DB BD 6. Refle Prop. of 7. ADB CBD 7. SAS Steps 5,, Statements Reasons 1. QPS TPR. RPS RPS. Refle. Prop. of 3. QPR TPS 3. Subst. Prop. of. PQ PT, PR PS. Given 5. PQR PTS 5. SAS Steps 3, 3. m FKJ + m KFJ + m FJK = = = 80 = 16 KJ = HJ = 7, so KJ HJ b the def. of. FJK FJH b the Rt. Thm. FJ FJ b the Refle. Prop. of. So FJK FJH b SAS. 35. m KFJ = m HFJ + 6 = = FK = FH = 171, so FK FH b the def. of. KFJ HFJ b the def. of bisector. FJ FJ b the Refle. Prop. of. So FJK FJH b SAS. TRiangle congruence: ASA, AAS, and HL CHECK IT OUT! 1. Yes; the is uniquel determined b AAS.. B the Alt. Int. Thm., KLN MNL. LN NL b the Refle. Prop. of. No other congruence relationships can be determined, so ASA cannot be applied. 3. JL bisects KLM. K M Given KLJ MLJ JKL JML AAS Given Def. of bisector JL JL Refle. Prop. of. Yes; it is given that AC DB. CB BC b the Refle. Prop. of. Since ABC and DCB are rt., ABC DCB b HL. THINK AND DISCUSS 1. No; the sides are not corr. sides.. Possible answer: corr. and sides 3. Words Pictures Words Pictures Eercises Def. of All 6 corr. parts of are. guided practice SSS 3 sides of 1 are to 3 sides of another. SAS sides and an included of 1 are to sides and an included in another. ASA AAS HL and an included side of 1 are to and included side in another. and a side of 1 are to their corr. parts in another. A leg and hp. of 1 rt. are to a leg and hp. in another rt. 1. The included side BC is enclosed between ABC and ACB ft A B C 3. Yes; the is determined b AAS.. Yes; b the Def. of bisector, TSV RSV and TVS RVS. SV SV b the Refle. Prop. of. So VRS VTS b ASA. 5. No; ou need to know that a pair of corr. sides are. 6a. QS SQ b. RQS PSQ c. Rt. Thm. d. AAS 7. Yes; it is given that D and B are rt. and AD BC. ABC and CDA are rt. b def. AC CA b the Refle. Prop. of. So ABC CDA b HL. 75 Holt McDougal Geometr

18 8. No; ou need to know that VX VZ. practice and Problem solving 9. X 37 6 km F 69 7 Y 10. Yes; the is uniquel determined b ASA. 11. No; ou need to know that MKJ MKL. 1. Yes; b the Alt. Int. Thm., SRT UTR, and STR URT. RT TR b the Refle. Prop. of. So RST TUR b ASA. 13a. A D b. Given c. C F d. AAS 1. No; ou need to know that K and H are rt Yes; E is a mdpt. So b def., BE CE, and AE DE. A and D are b the Rt. Thm. B def., ABE and DCE are rt.. So ABE DCE b HL. 16. AAS proves ADB CDB; reflection 17. FEG QSR; rotation a. No; there is not enough information given to use an of the congruence theorems. b. HL can be used, since also JL JL. 0. Proof B is incorrect. The corr. sides are not in the correct order. 1. A D C B F E It is given that ABC and DEF are rt.. AC DF, BC EF, and C and F are rt.. C F b the Rt. Thm. Thus ABC DEF b SAS.. Statements Reasons 1. AD ǁ BC. DAE BCE. Alt. Int. Thm. 3. AED CEB 3. Vert. Thm.. AD CB. Given 5. AED CEB 5. AAS Steps, 3, 3. Statements Reasons 1. KM JL. JKM and LKM are rt.. Def. of 3. JKM LKM 3. Rt. Thm.. JM LM, JMK LMK. Given 5. JKM LKM 5. AAS Steps 3,. Since sides and the included are equal in measure and therefore, ou could prove the using SAS. You could also use HL since the are rt.. 5. Check students constructions. test prep 6. A Need XVZ XWY for ASA. 7. J From figure, corr. side pairs and included pair are, i.e., SAS. 8. C Alt. Int. Thm. gives two pairs, and one nonincluded side pair is given. AAS proves AED CEB. 9. G For AAS, need RT UW. So: RT = UW 6 - = + 7 = 9 = No; check students drawings and constructions; since the lengths of the corr. sides of the are not equal, the are not even if the corr. have the same measure. challenge and etend 31. Yes; the sum of the measures in each must be 180, which makes it possible to solve for and. The value of is 15, and the value of is 1. Each has measuring 8, 68, and 30. VU VU b the Refle. Prop. of. So VSU VTU b ASA or AAS. 3. Statements Reasons 1. ABC is equil.. AC BC. Def. of equil. 3. C is mdpt. of DE. 3. Given. DC EC. Def. of mdpt. 5. DAC and EBC are. and supp. 6. DAC and EBC are rt.. 7. DAC and EBC are rt.. 5. Given 6. that are and supp. are rt.. 7. Def. of rt. 8. DAC EBC 8. HL Steps, 76 Holt McDougal Geometr

19 33. A D B C E F Case 1: Given rt. ABC and rt. DEF with A D and AB DE Statements 1. A D. AB DE. Given A Reasons 3. B E 3. Rt. Thm.. ABC DEF. ASA Steps 1,, 3 D B C E F Case ; given rt. ABC and rt. DEF with A D and BC EF Statements 1. A D. BC EF. Given Reasons 3. B E 3. Rt. Thm.. ABC DEF. ASA Steps 1, 3, 3. Third Thm.; if the third pair is, then the are also b AAS Triangle Congruence: CPCTC CHECK IT OUT! 1. JL = NL and KL = ML, so JL NL and KL ML. B Vert. Thm., MLN KLJ. B SAS, MLN KLJ. B CPCTC, JK = NM = 1 ft. PR bisects QPS and QRS. Given QPR SPR QRP SRP Def. of bisector PQR PSR ASA PR PR Refle. Prop. of PQ PS CPCTC 3. Statements Reasons 1. J is mdpt. of KM and NL.. KJ MJ and LJ NJ. Def. of mdpt. 3. KJL MJN 3. Vert. Thm.. KJL MJN. SAS Steps, 3 5. LKJ NMJ or JLK JNM 6. KL ǁ MN 5. CPCTC 6. Conv. of Alt. Int. Thm.. Use Distance Formula to find side lengths. JK = ÇÇÇÇÇÇÇÇÇÇÇÇ ( - (-1) ) + ((-1) - (-) ) = ÇÇÇ = ÇÇ 10 KL = ÇÇÇÇÇÇÇÇÇÇÇ ((-) - ) + (0 - (-1) ) = ÇÇÇ = ÇÇ 17 JL = ÇÇÇÇÇÇÇÇÇÇÇÇ ((-) - (-1) ) + (0 - (-) ) = ÇÇÇ 1 + = Ç 5 RS = ÇÇÇÇÇÇÇÇ (5 - ) + ( - 3 ) = ÇÇÇ = 10 ÇÇ ST = ÇÇÇÇÇÇÇÇ (1-5 ) + (1 - ) = ÇÇÇ = ÇÇ 17 RT = ÇÇÇÇÇÇÇÇ (1 - ) + (1-3 ) = ÇÇÇ 1 + = Ç 5 So JK RS, KL ST, and JL RT. Therefore, JKL RST b SSS, and JKL RST b CPCTC. THINK AND DISCUSS 1. SAS; UW XZ ; U X; W Z. A D B E C F Eercises ABC DEF CPCTC Guided Practice AB DE BC EF AC DF 1. Corr. and corr. sides. BCA DCE b Vert. Thm, CBA CDE b Rt. Thm., and BC DC (given). Therefore ABC EDC b ASA. B CPCTC, AB DE, so AB = DE = 6.3 m. 3a. Def. of b. Rt. Thm. c. Refle. Prop. of d. Def. of mdpt. e. RXS RXT f. CPCTC. Statements Reasons 1. AC AD, CB DB. AB AB. Refle. Prop. of 3. ACB ADB 3. SSS Steps 1,. CAB DAB. CPCTC 5. AB bisects CAD. 5. Def. of bisector 77 Holt McDougal Geometr

20 5. Use Distance Formula to find side lengths. EF = ÇÇÇÇÇÇÇÇÇÇÇ ((-1) - (-3) ) + (3-3 ) = ÇÇÇ + 0 = FG = ÇÇÇÇÇÇÇÇÇÇÇ ((-) - (-3) ) + (0-3 ) = ÇÇÇ = ÇÇ 10 EG = ÇÇÇÇÇÇÇÇÇÇÇ ((-) - (-1) ) + (0-3 ) = ÇÇÇ = ÇÇ 10 JK = ÇÇÇÇÇÇÇÇÇÇÇ (0 - ) + ((-1) - (-1) ) = ÇÇÇ + 0 = KL = ÇÇÇÇÇÇÇÇÇ (1 - ) + ( - (-1) ) = ÇÇÇ = 10 ÇÇ JL = ÇÇÇÇÇÇÇÇÇ (1-0 ) + ( - (-1) ) = ÇÇÇ = ÇÇ 10 So EF JK, FG KL, and EG JL. Therefore EFG JKL b SSS, and EFG JKL b CPCTC. 6. Use Distance Formula to find side lengths. AB = ÇÇÇÇÇÇÇÇ ( - ) + (1-3) = ÇÇÇ + = Ç BC = ÇÇÇÇÇÇÇÇÇ (1 - ) + ((-1) - 1) = ÇÇÇ 9 + = ÇÇ 13 AC = (1 ÇÇÇÇÇÇÇÇÇ - ) + ((-1) - 3) = ÇÇÇ = ÇÇ 17 RS = ÇÇÇÇÇÇÇÇÇÇÇÇ ((-3) - (-1) ) + ((-) - 0 ) = ÇÇÇ + = Ç ST = ÇÇÇÇÇÇÇÇÇÇÇÇ (0 - (-3) ) + ((-) - (-) ) = ÇÇÇ 9 + = ÇÇ 13 RT = ÇÇÇÇÇÇÇÇÇÇÇ (0 - (-1) ) + ((-) - 0 ) = ÇÇÇ = ÇÇ 17 So AB RS, BC ST, and AC RT. Therefore ABC RST b SSS, and ACB RTS b CPCTC. Practice and Problem Solving 7. ABC EDC b Rt. Thm., ACB ECD b Vert. Thm., and BC DC. So ABC EDC b ASA. B CPCTC, AB = DE = 0 ft. 8. Statements Reasons 1. M is mdpt. of PQ and RS.. PM QM, RM SM. Def. of mdpt. 3. PMS QMR 3. Vert. Thm.. PMS QMR. SAS Steps, 3 5. QR PS 5. CPCTC 9. Statements Reasons 1. WX XY YZ ZW. ZX XZ. Refle. Prop. of 3. WXZ YZX 3. SSS, steps 1,. W Y. CPCTC 10. Statements Reasons 1. G is mdpt. of FH.. FG = HG. Def. of mdpt. 3. FG HG 3. Def. of.. Draw EG. 5. EG EG 6. EF EH. Eactl 1 line through an pts. 5. Refle. Prop. of 6. Given 7. EGF EGH 7. SSS Steps 3, 5, 6 8. EFG EHG 8. CPCTC Supp. Thm. 11. Statements Reasons 1. LM bisects JLK.. JLM KLM. Def. of bisector 3. JL KL 3. Given. LM LM. Refle. Prop. of 5. JLM KLM 5. SAS Steps 3,, 6. JM KM 6. CPCTC 7. M is mdpt. of JK. 7. Def. of mdpt. 1. RS = ÇÇÇÇÇÇÇÇ ( - 0 ) + ( - 0 ) = ÇÇÇ + 16 = 5 Ç ST = ÇÇÇÇÇÇÇÇÇ ((-1) - ) + ( - 3 ) = ÇÇÇ = 10 ÇÇ RT = ÇÇÇÇÇÇÇÇÇ ((-1) - 0 ) + (3-0 ) = ÇÇÇ = 10 ÇÇ UV = ÇÇÇÇÇÇÇÇÇÇÇÇ ((-3) - (-1) ) + (( - ) - 0 ) = ÇÇÇ + 16 = Ç 5 VW = ((-) ÇÇÇÇÇÇÇÇÇÇÇÇÇ - (-3) ) + ((-1) - (-) ) = ÇÇÇ = ÇÇ 10 UW = ((-) ÇÇÇÇÇÇÇÇÇÇÇÇ - (-1) ) + ((-1) - 0 ) = ÇÇÇ = ÇÇ 10 So RS UV, ST VW, and RT UW. Therefore, RST UVW b SSS, and RST UVW b CPCTC. 78 Holt McDougal Geometr

21 13. AB = ÇÇÇÇÇÇÇÇÇ ( - (-1) ) + (3-1 ) = ÇÇÇ 9 + = 13 ÇÇ BC = ( ÇÇÇÇÇÇÇÇÇ - ) + ((-) - 3 ) = ÇÇÇ = 5 AC = ÇÇÇÇÇÇÇÇÇÇÇ ( - (-1) ) + ((-) - 1 ) = ÇÇÇ = 3 Ç DE = ÇÇÇÇÇÇÇÇÇÇÇÇ ((-1) - ) + ((-5) - (-3) ) = ÇÇÇ 9 + = ÇÇ 13 EF = ((-1) ÇÇÇÇÇÇÇÇÇÇÇÇ - (-1) ) + (0 - (-5) ) = ÇÇÇ = 5 DF = ÇÇÇÇÇÇÇÇÇÇÇ ((-1) - ) + (0 - (-3) ) = ÇÇÇ = 3 Ç So AB DE, BC EF, and CA DF. Therefore, ABC DEF b SSS, and BAC EDF b CPCTC. 1. Statements Reasons 1. QRS is adj. to QTS. QS bisects RQT. R T.. RQS TQS. Def. of bisector 3. QS QS 3. Refle. Prop. of. RSQ TSQ. AAS Steps 1,, 3 5. RS TS 5. CPCTC 6. QS bisects RT. 6. Def. of bisector 15. Statements Reasons 1. E is the mdpt. of AC and BD.. AE CE, BE DE. Def. of mdpt. 3. AEB CED 3. Vert Thm.. AEB CED. SAS Steps, 3 5. A C 5. CPCTC 6. AB ǁ CD 6. Conv. of Alt. Int. Thm. 16a. ADB, ADC are rt., hp. lengths are =, corr. leg lengths are =. So HL proves ADB ADC. b. Statements Reasons 1. AD BC. ADB and ADC are rt.. 3. ADB and ADC are rt.. Def. of 3. Def. of rt.. AB = AC = 0 in.. Given 5. AB AC 5. Def. of 6. AD AD 6. Refle. Prop. of 7. ADB ADC 7. HL Steps 5, 6 8. BD CD 8. CPCTC c. BD + AD = AB BD + 10 = 0 BD = ÇÇÇÇÇ in. BC = BD 3.6 in. 17. are b SAS = = 18. are b ASA. + 1 = 6-1 = = Statements Reasons 1. PS = RQ. PS RQ. Def. of 3. m 1 = m 3. Given. 1. Def. of 5. SQ QS 5. Refle. Prop. of 6. PSQ RQS 6. SAS Steps,, CPCTC 8. m 3 = m 8. Def. of 0. Statements Reasons 1. m 1 = m, m 3 = m. 1, 3. Def. of 3. SQ SQ 3. Refle. Prop. of. PSQ RSQ. ASA Steps, 3 5. PS RS 5. CPCTC 6. PS = RS 6. Def. of 1. Statements Reasons 1. PS = RQ, PQ = RS. PS RQ, PQ RS. Def. of 3. SQ QS 3. Refle. Prop. of. PSQ RQS. SSS Steps, CPCTC 6. PQ ǁ RS 6. Conv. of Alt. Int. Thm.. Yes; JKM LMK b SSS, so JKM LMK b CPCTC. Therefore, JK ǁ ML b Conv. of Alt. Int. Thm. 3. A B C D The segs. CA, CD, and CB must be. ACB DCB. If ACB DCB b SAS, then AB = DB. 79 Holt McDougal Geometr

22 Test prep. C Onl wa to get a second pair is first to prove are and then to use CPCTC. But ou would use CPCTC to prove AC AD directl. 5. G LNK NLM, so b CPCTC LNK NLM. 6. C 6 = + 5_ 5 = 5_ = 1_ 10 + = 0 = 0-10 = _ = G Onl corr. parts are ever used., ǁ lines, lines all are used. 8. B RS = ÇÇÇÇÇÇÇÇ (3 - ) + (3-6 ) = 10 ÇÇ ST = ÇÇÇÇÇÇÇÇ ( - 6 ) + (6-6 ) = RT = ÇÇÇÇÇÇÇÇ (6-3 ) + (6-3 ) = 3 Ç These lengths onl match the coordinates in B. Challenge and Etend 9. An diagonal on an face of the cube is the hp. of a rt. whose legs are edges of the cube. An of these are b SAS (or LL). Therefore, an diagonals are b CPCTC. 30. Statements Reasons 1. Draw MK. 1. Through an pts. there is eactl 1 line.. MK KM. Refle. Prop. of 3. JK LM, JM LK 3. Given. JKM LMK. SSS Steps, 3 5. J L 6. CPCTC 31. Statements Reasons 1. R is the mdpt. of AB.. AR BR. Def. of mdpt. 3. RS AB 3. Given. ARS and BRS. Def. of are rt. 5. ARS BRS 5. Rt. Thm. 6. RS RS 6. Refle. Prop. of 7. ARS BRS 7. SAS Steps, 5, 6 8. AS BS 8. CPCTC 9. ASD BSC 9. Given 10. S is the mdpt. of DC. 10. Given 11. DS = CS 11. Def. of mdpt. 1. ASD BSC 1. SAS Steps 8, 9, A E (given), B and D are rt. (from figure), and BC CD (from figure). Therefore, ABC EDC b HL. B CPCTC, AB = DE. B Pthag. Thm., CD + DE = CE DE = 1-10 AB = DE = ÇÇÇÇÇ ft -8-8 introduction to Coordinate Proof CHECK IT OUT! 1. You can place the longer leg along the -ais and the other leg along the -ais Proof: ABC is a rt. with height AB and base BC. The area of ABC = 1_ bh = 1_ ()(6) = 1 square units B Mdpt. Formula, coordinates of D = ( _ 0 +, _ ) = (, 3). The -coord. of D is height of ADB, and base is 6 units. The area of ADB = 1_ bh = 1_ ()(6) = 6 square units Since 6 = 1_ (1), area of ADB is 1 area of ABC. 3. Possible answer: (0, p) (p, p) (0, 0) (p, 0). ABC is a rt. with height j and base n. The area of ABC = 1_ bh = 1_ (n)(j) = nj square units B the Mdpt. Forumla, the coords. of D are (n, j). The base of ABC is j units and the height is n units. So the area of ADB = 1_ bh = 1_ (j)(n) = nj square units Since nj = 1_ (nj), the area of ADB is 1 the area of ABC. THINK AND DISCUSS 1. Possible answer: B using variables, our results are not limited to specific numerical values. 80 Holt McDougal Geometr

23 . Possible answer: The wa ou position the figure will affect the coords. assigned to the vertices and therefore, our calculations. 3. Possible answer: If ou need to calculate the coords. of a mdpt., assigning p allows ou to avoid using fractions.. Use origin as a verte. Center figure at origin. Center side of figure at origin. Eercises Guided Practice Use aes as sides of figure. 1. Possible answer: In coordinate geometr, a coord. proof is one in which ou position figures in the coord. plane to prove a result B the Mdpt. Forumla, the coords of A are (0, 3) and the coords. of B are (, 0). B the Dist. Formula, PQ = ÇÇÇÇÇÇÇÇ (0-8) + (6-0) = ÇÇÇÇÇ (-8) + 6 = 6 ÇÇÇÇ + 36 = ÇÇ 100 = 10 units. AB = ÇÇÇÇÇÇÇÇ (0 - ) + (3-0) = ÇÇÇÇÇ (-) + 3 = ÇÇÇ = ÇÇ 5 = 5 units. So AB = 1_ PQ. 5. Possible answer: (0, m) (0, 0) (n, 0) 6. Possible answer: (0, b) (a, b) 7. (0, 0) P A R (0, 0) (0, a) B (a, 0) (b, 0) Q B the Mdpt. Formula, the coords. of A are (0, a) and the coords of B are (b, 0). B the Dist. Formula, PQ = ÇÇÇÇÇÇÇ (0 - b) + (a) AB = ÇÇÇÇÇÇÇÇ (0 - b ) + (a - 0 ) = ÇÇÇÇÇÇ (-b) + (a) = ÇÇÇÇ b + a = ÇÇÇÇ b + a units So AB = 1_ PQ. Practice and Problem Solving 8. Possible answer: - 9. Possible answer: 10. B 0 8 E 0 C F = ÇÇÇÇÇ (-b ) + a = ÇÇÇÇ b + a units A D E = ( _ 0 + 0, _ ) = (0, 5) F = ( _ 6 + 6, _ ) = (6, 5) BC = ÇÇÇÇÇÇÇÇÇ (6-0 ) + (10-10 ) = 6 units. EF = ÇÇÇÇÇÇÇÇ (6-0 ) + (5-5 ) = 6 units. So EF = BC. 81 Holt McDougal Geometr

24 11. Possible answer: (0, m) (m, m) (0, 0) (m, 0) 1. Possible answer: 13. (0, ) (3, ) (0, 0) A D (0, a) (c, a) E B (0, 0) C F (c, 0) (3, 0) B the Mdpt. Formula, the coords. of E are (0, a) and the coords of F are (c, a). B the Dist. Formula, AD = ÇÇÇÇÇÇÇÇÇ (c - 0) + (a - a) = ÇÇ (c) = c units. Simlarl, EF = ÇÇÇÇÇÇÇÇ (c - 0) + (a - a) = ÇÇ (c) = c units. So EF = AD. 1. Let endpts. be (, ) and (z, w). B Mdpt. Formula, (0, 0) = ( _ + z, _ + w ) _ + z = 0 + z = 0 z = - _ + w = 0 + w = 0 w = - Endpts are (, ) and (-, -). 15a. 0 6 b. Total distance = EW + WC = ÇÇÇÇÇÇÇÇ (3-0 ) + (3-0 ) + ÇÇÇÇÇÇÇÇ (6-3 ) + (0-3 ) = 3 Ç + 3 Ç = 6 Ç Let A = (0, 0), B = (a, 0), and C = (0, a). Perimeter = AB + BC + CA = a + ÇÇÇÇÇÇÇÇ (0 - a ) + (a - 0 ) + a = a (3 + Ç 5 ) units ABC has base AB, height AC. Area = 1_ bh = 1_ ( a)(a) = a square units 17. Let A = (0, 0), B = (s, 0), C = (s, t), and D = (0, t). Perimeter = AB + BC + CD + DA = s + t + s + t = s + t units Area = lw = st square units 18. (n, n) 19. (p, 0) 0. ÇÇÇÇÇÇÇÇÇÇÇÇÇÇ (-3. - (-5) ) + ( ) 1.8 units ÇÇÇÇÇÇÇÇÇÇÇÇÇÇ (- - (-3.) ) + ( ) 0.9 units ÇÇÇÇÇÇÇÇÇÇÇÇÇ (- - (-5) ) + ( ) 1.1 units 1.8 is twice 0.9. The dist. between of the locations is appro. twice the dist. between another locations. 1. AB = ÇÇÇÇÇÇÇÇÇÇÇÇ (70 - (-30) ) + ((-30) - 50 ) 18 nautical miles Mdpt. of AB = ( _, _ 50 + (-30) ) = (0,10) So, P is the mdpt of AB.. S 0 P T Q R The area of the rect. is A = lw = 3() = 6 square units. For RST, the base is 3 units, and the height is 1 unit. So the area of RST = 1_ bh = 1_ (3)(1) = 1.5 square units. Since 1_ (6) = 1.5, the area of RST is 1_ of the area of the rect. 3. B Dist. Formula, AB = ÇÇÇÇÇÇÇÇÇ ( - 1 ) + ( - 1 ) and AM = ÇÇÇÇÇÇÇÇÇÇÇÇÇÇ (_ ) + ( _ ) ÇÇÇÇÇÇÇÇÇÇÇÇÇÇÇ ( = _ _ 1 ) + ( _ _ ) 1_ ( - 1 ) + 1_ = ÇÇÇÇÇÇÇÇÇÇÇ ( - 1 ) = 1_ ÇÇÇÇÇÇÇÇÇ ( - 1 ) + ( - 1 ) So AM = 1_ AB. 8 Holt McDougal Geometr

25 . L M K P - 0 Proof: B Dist. Formula, KL = ÇÇÇÇÇÇÇÇÇ (- + ) + (1-3 ) = ÇÇÇ 0 + = MP = ÇÇÇÇÇÇÇÇ (1-1 ) + (3-1 ) = ÇÇÇ 0 + = LM = ÇÇÇÇÇÇÇÇÇ (- - 1 ) + (3-3 ) = ÇÇÇ = 3 PK = ÇÇÇÇÇÇÇÇ (1 + ) + (1-1 ) = ÇÇÇ = 3 Thus KL = MP and LM = PK b Trans. Prop. of. KL MP and LM PK b def. of, and KM MK b Refle. Prop. of. Thus KLM MPK b SSS. 5. You are assuming the figure has a rt.. 6a. BD = BC + CD = AE + CD = = 38 in. B Dist. Formula, DE = ÇÇÇÇÇ CD + CE CE = DE - CD CE = ÇÇÇÇ 6-10 = in. b. B = (, 0); C = (, 8); D = (, 38); E = (0, 8) Test Prep 7. B; Mdpt. Formula shows B is true. 8. F; G, H, and J are all possible vertices. 9. D; Perimeter = a + b + a + b = a + b 30. H; ( _-1 + 7, _ + 8 ) = (3, 5) = C Challenge and Etend 31. (a + c, b) 3. (n + p n, h h) = (p, 0) 33. Possible answer: Rotate 180 about (0, 0) and translate b (0, s). The new coords. would be (0, 0), (s, 0), (0, s). 3. E is intersection of given lines. At E, = _ g and f = - g + g. f _ g + g Set eqns. = to each other. f = - g f _ g = g f = f = _ g f = _ g f f = g E = (f, g) Combine like terms. Simplif. Given Subst. Simplif isosceles and Equilatral Triangles CHECK IT OUT! ; since there are 6 months between September and March, the measures will be appro. the same between Earth and the star. B the Conv. of the Isosc. Thm., the created are isosc., and the dist. is the same. a. m G = m H = m F + m G + m H = = 180 = 13 = 66 Thus m H = = 66. b. m N = m P 6 = = = 8 Thus m N = 6 = 6(8) = JKL is equilateral. t - 8 = t + 1 t = 9 t =.5 JL = t + 1 = (.5) + 1 = 10. Proof: B Mdpt. Formula, coords. of X are ( -a + 0, 0 + b ) = (-a, b), coords. of Y are ( a + 0, 0 + b ) = (a, b), and coords of Z are ( -a + a, ) = (0, 0). B Dist. Formula, XZ = ÇÇÇÇÇÇÇÇ (0 + a ) + (0 - b ) = ÇÇÇÇ a + b, and YZ = ÇÇÇÇÇÇÇÇ (0 - a ) + (0 - b ) = ÇÇÇÇ a + b Since XZ = YZ, XZ YZ b definition. So XYZ is isosc. THINK AND DISCUSS 1. An equil. is also an equiangular, so the 3 have the same measure. The must add up to 180 b the Sum Thm. So each must measure 60.. Triangle Equilateral: Equiangular: 83 Holt McDougal Geometr