Geometry Student Text and Homework Helper

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1 Topic Lesson - pp m vr. Smple: plne EG, plne FG 3. E,, F, G 5. RS >, SR >, ST >, TS >, TW >, WT >, TR >, RT >, WR >, RW >, WS >, SW > 7. < VW > 9. plne TSR, plne TSW. plne VWX, plne VWS 3. X W 5. X W U T V S Q R Q R 7. noncoplnr 9. m vr. Smple: You cn represent plne b drwing four-sided shpe. However, plnes do not hve boundries, so the etend pst the drwn edges without end. Since plne does not hve n endpoint, the intersection of the two plnes our friend drew cnnot be point.. Not lws; > contins >, but the re not the sme r. 3. lws 5. sometimes 7. U T Loction of cell phone ostulte -, the loction of the cell phone nd point determine line, nd the loction of the cell phone nd point determine line. ostulte -, the two lines intersect t, or shre, onl one point. Since the cell phone signl is on both lines, its loction must be t the intersection of the lines. 9. m vr. Smple: now, think, eist O O es F Lesson - pp bout h, min 7. 9 b. Y = 9, XY = 8 9. = 5; =, D =. Not lws; the Segment ddition ostulte cn be used onl if, Q, nd R re colliner points. no V S or no 9. es. The distnce or 5 mi. The driver dded the vlues insted of subtrcting them or G Lesson -3 pp. 9. XYZ, ZYX, Y 3. JKM, MKJ, or 5. 90, right 7. m vr. Smple: R 9. FHG es; S bout bout 90 ; right 5. ngle ddition ostulte 7. = 8; m O = 5, m OD = m RQS = 3, m TQS = J Lesson - pp Yes, the ngles shre common side nd verte, nd hve no interior points in common. 3. No, the re supplementr. 5. EO 7. m vr. Smple: O, DO 9. 35, No; the do not hve common verte. 5. m EFG = 69, m GFH = 7. D; b. 8 c. 9; No; J nd D re not mrked s. 3. Yes; the re formed b < JF > nd < ED >. 5. Lesson -5 pp m vr. Smple: X V V Z Y V V Find segment on < XY > so tht ou cn construct < YZ > s its perpendiculr bisector. X T Y Z

2 3. m vr. Smple: oth constructions involve drwing rcs with the sme rdius from two different points, nd using the point(s) of intersection of those rcs. rcs must intersect t two points for the # bis., but onl one point for the bis F Q R b. The three ngle bisectors meet t point. m. 3. no 5. D 7. - = - - = 0 ( - )( + ) = 0 = or = - (not possible) = c. For n tringle, the three ngle bisectors meet t point. 7. Not possible; the cm sides meet on the cm side, so the do not form tringle. 9. segment hs ectl one midpoint; using the Ruler ostulte (ost. -5), ech point corresponds with ectl one number, nd ectl one number represents hlf the length of segment. b. segment hs infinitel mn bisectors becuse infinitel mn lines cn be drwn through the midpoint. c. In the plne with the segment, there is one # bis. becuse onl one line in tht plne cn be drwn through the midpoint so tht it forms right ngle with the given segment. d. onsider the plne tht is the # bis. of the segment. n line in tht plne tht contins the midpoint of the segment is # bis. of the segment, nd there re infinitel mn such lines. b. O Technolog Lb -5 pp Yes. It is possible to mke m HGF = 90 nd EG = GF. When these conditions re met, < HG > is perpendiculr bisector. 3. The position of EF reltive to < GH > cn chnge, wheres the position of reltive to < D > is fied. 5. sk our techer to check our work. b. no ctivit Lb -5 p. 36. es 3. no; not plne figure 5. Smple: FWMX; sides re F, W, WM, MX, XF; ngles re F,, W, M, X 7. Smple: GNHET; sides re G, GN, NH, HE, E, T, T; ngles re, G, N, H, E,, T 9. nongon or ennegon, conve Topic Review pp ngle bisector 3. construction 5. < QR > 7. True; ostulte - sttes, Through n two points, there is ectl one line , cute m vr. Smple: D nd D 9. m vr. Smple: D nd EDF c. O is the center of the circle. 3. V W 5. L M

3 TEKS umultive rctice pp. 0. D D No, is, Q, nd R re not colliner, then Q is not on R, so it cnnot be the midpoint of R.. + 6; Smple nswer: m O = m OD = + since O is the ngle bisector of OD. m O = m O since O is the ngle bisector of OD. Thus, m O = ( + ) = + 6. b. =. Smple nswer: m OD = m OD since O is the ngle bisector of O. Thus, + = (0) = 60. If + = 60, then =. Topic Lesson - pp Double the previous term; 80, dd -, +3, -, +5, c; -3,. 5. The numertor is, nd the denomintor is the net whole number; 5, the first letters of the counting numbers; N, T 9. Multipl the previous number b, b 3, b, b 5, c; 70, U.S. coins of descending vlue; dime, nickel 3. zodic signs; Gemini, ncer cm 9. blue. 75 F 3. nd 5. m vr. Smples re given. 3. two right ngles 5. - nd sì-shí-sān; lìu-shí-qī; bā-shí-sì b. Yes; the second prt of the number repets ech ten numbers. 9. ird ount Number of Species Yer b. m vr. Smple: Using just the dt from 005 to 008, the gin is 7 species in 3 ers, or between nd 3 species ech er. The er 05 is 7 ers fter 008, so the number of new species will be between nd more thn 90; n estimte is or 07 species. 3. * : 6 squres; * : 9 squres; 3 * 3 : 36 squres; * : 5 squres; 5 * 5 : 6 squres; 6 * 6 : 9 squres; 7 * 7 : squres; 8 * 8 : squre; totl number of squres: Lesson - pp Hpothesis: You re n mericn citizen. onclusion: You hve the right to vote. 3. Hpothesis: You wnt to be helth. onclusion: You should et vegetbles. 5. If ou hve never mde mistke, then ou hve never tried nthing new. 7. Yes, he is correct; both re true, becuse conditionl nd its contrpositive hve the sme truth vlue. 9. If point is in the first qudrnt of coordinte plne, then both coordintes of tht point re positive.. If number is whole number, then it is n integer. 3. flse; Meico 5. true 7. m vr. Smple: If person is pitcher, then tht person is bsebll pler. If person is bsebll pler, then tht person is n thlete. If person is pitcher, then tht person is n thlete. 9. Obtuse ngles 00. eople who wnt to help others ece orps volunteers 3. If - is positive, then is negtive; true. 5. If 7 0, then 6 0; flse: counteremple is =. 7. onditionl: If person is pinist, then tht person is musicin. onverse: If person is musicin, then tht person is pinist. Inverse: If person is not pinist, then tht person is not musicin. ontrpositive: If person is not musicin, then tht person is not pinist. The conditionl nd the contrpositive re true. The converse nd the inverse re flse; counteremple: percussionist is musicin. 9. onditionl: If number is n odd nturl number less thn 8, then the number is prime. onverse: If number is prime, then it is n odd nturl number less thn 8. Inverse: If number is not n odd nturl number less thn 8, then the number is not prime. ontrpositive: If number is not prime, then it is not n odd nturl number less thn 8. ll four sttements re flse; counteremples: nd. 3. If ou wer Snzz snekers, then ou will look cool. 33. If two figures re congruent, then the hve equl res. 35. ll integers divisible b 8 re divisible b. 37. Some musicins re students. 39. J Lesson -3 pp onverse: If two segments re congruent, then the hve the sme length; true. iconditionl: Two segments hve the sme length if nd onl if the re congruent. 3. onverse: If it is Independence D in the United Sttes, then it is Jul ; true. iconditionl: In the United Sttes, it is Jul if nd onl if it is Independence D. 5. Yes; it uses clerl understood terms, is precise, nd is reversible. 7. Tht sttement, s biconditionl, is n ngle is right ngle if nd onl if it is greter thn n cute ngle. ounteremples to tht sttement re obtuse ngles nd stright ngles. 9. point is in Qudrnt III if nd onl if it hs two negtive coordintes. 3

4 . number is whole number if nd onl if it is nonnegtive integer. 3. If n integer is divisible b 00, then its lst two digits re zeros. If the lst two digits of n integer re zeros, then the integer is divisible b If =, then = or = -. If = or = -, then =. 7. good definition 9. If nd re liner pir, then nd re supplementr.. If nd re liner pir, then nd re djcent nd supplementr ngles. 3. m vr. Smple: line is circle on the sphere formed b the intersection of the sphere nd plne contining the center of the sphere If ou go to the store, then ou wnt to bu milk; flse. b. m vr. Smple: counteremple is going to the store becuse ou wnt to bu juice. Lesson - pp No conclusion is possible; the conclusion hs been stisfied, but the hpothesis hs not been stisfied. 3. No conclusion is possible; the sme sttement does not pper s the conclusion of one conditionl nd s the hpothesis of the other conditionl. 5. $5.99; our fmil goes to our fvorite resturnt on the night of our weekl gme, which is on Tuesd. hicken fingers re $5.99 on Tuesd. 7. Must be true; b E nd, it is brekfst time; b, urtis is drinking wter. 9. Is not true; b E nd, it is brekfst time; b, urtis drinks wter nd nothing else.. Is not true; b nd E, it is brekfst time; b D, Julio is drinking juice nd nothing else. 3. lsk s Mount McKinle is the highest mountin in the U.S. 5. From the tble, if qurk hs rest energ 50 MeV nd chrge of - 3 e, then the flvor of the qurk is strnge; given qurk hs rest energ 50 MeV nd chrge of - 3 e. the Lw of Detchment, the flvor of the given qurk is strnge. 7. Lesson -5 pp Mult. rop. of Eq. b. Distr. rop. c. dd. rop. of Eq. 3. Seg. dd. ost. b. Subst. rop. c. Distr. rop. d. Distr. rop. e. Subtr. rop. of Eq. f. Div. rop. of Eq. 5. Since LR nd RL re two ws to nme the sme segment nd nd re two ws to nme the sme, then both sttements re emples of sing tht something is to itself. 7. KM = 35 (Given); KL + LM = KM (Seg. dd. ost.); ( - 5) + = 35 (Subst. rop.); - 5 = 35 (Distr. rop.); = 0 (dd. rop. of Eq.); = 0 (Div. rop. of Eq.); KL = - 5 (Given); KL = (0) - 5 (Subst. rop.); KL = 5(Simplif). 9. The error is in the 5th step when both sides of the eqution re divided b b -, which is 0, nd division b 0 is not defined.. Sm. rop. of 3. Given b. midpt. divides seg. into two segments. c. Substitution rop. of Eq. d. = e. Div. rop. of Eq. 5. K Trnsitive onl; cnnot be tller thn ; if is tller thn, then is not tller thn Lesson -6 pp = 5, + 0 = 5, - 35 = 5 3. Vert. s Thm. b. 6 c. Vert. s Thm. d. Trns. rop. of nd 3 re given s vert. s. ecuse nd form liner pir, nd re suppl. ecuse nd 3 form liner pir, nd 3 re suppl. So m + m = 80 nd m + m 3 = 80 b the def of suppl. s. the Trns. rop. of Eq., m + m = m + m 3. Subtrct m from both sides. the Subtr. rop. of Eq., m = m 3. s with equl mesure re, so , 5, 5. it is given b. m V c. 80 d. Division e. right 3. Theorem -5: If two ngles re nd suppl., then ech is right ngle. 5. m = 30, m = m = 90, m = Smple nswer: The intersecting lines form two pirs of verticl ngles. the Verticl ngles Theorem, becuse one ngle mesures 90, its verticl ngle lso mesures 90. Ech of the remining two ngles forms liner pir with 90 ngle, so ech hs mesure of 80-90, or 90. So ll four ngles re right ngles.. theorem; This is n emple of the ongruent omplements Theorem. 3. Y b. right c. m Y d. X Y 5. = 30, = 90; 60, 0, = 50, = 0; 80, 00, _ 7 b the Verticl ngles Theorem. 5 _ 8 b the Trnsitive ropert of ongruence. 8 _ 6 b the Verticl ngles Theorem. 5 _ 6 b the Trnsitive ropert of ongruence Topic Review pp conclusion 3. truth vlue 5. biconditionl 7. Divide the previous term b 0;, Subtrct 7 from the previous term; 6, -. m vr. Smple: - # = - nd - is not greter thn. 3. If person is motorcclist, then tht person wers helmet. 5. If two ngles form liner pir, then the ngles re supplementr. 7. onverse: If the mesure of n ngle is greter thn 90 nd less thn 80, then the ngle is obtuse. Inverse: If n ngle is not obtuse, then it is not true tht its mesure is greter thn 90 nd less thn 80. ontrpositive: If it is not true tht the mesure of ngle is greter thn 90 nd less thn 80, then the ngle is not obtuse. ll four sttements re true. 9. onverse: If ou pl n instrument, then ou pl the tub. Inverse: If ou do not pl the tub, then ou do not pl n instrument. ontrpositive: If ou do not pl n instrument, then ou do not pl the tub. The conditionl nd the contrpositive re true. The converse nd inverse re flse.. No; it is not reversible; mgzine is counteremple. 3. No; it is not reversible; line is counteremple.

5 5. If two ngles re complementr, then the sum of their mesures is 90; if the sum of the mesures of two ngles is 90, then the ngles re complementr. 7. m + m = If our fther bus new grdening gloves, then he will plnt tomtoes. 3. Y is compl. to, 3 is compl. to, nd re ll given. m = m b the def. of. nd re compl. b the Subst. rop. 3 b the ompl. Thm. TEKS umultive rctice pp D D D onverse: If ou live in the United Sttes, then ou live in Oregon; flse. Inverse: If ou do not live in Oregon, then ou do not live in the United Sttes; flse. ontrpositive: If ou do not live in the United Sttes, then ou do not live in Oregon; true. 7. Ones lce ower Digit b. 9; the ones digit hs four repeting digits (7, 9, 3, ). Since 3 divided b hs reminder of, the ones digit in 7 3 is the sme s the ones digit in 7. Topic 3 Lesson 3- pp lne JD nd plne ELH 3. < G >, < JE >, < L >, < F > 5. < G >, < DH >, < L > 7. 7 nd 6 (lines nd b with trnsversl d); nd 5 (lines b nd c with trnsversl e) 9. 5 nd 6 (lines d nd e with trnsversl b); nd (lines b nd e with trnsversl c); nd 3 (lines nd d with trnsversl c); 7 nd 3 (lines nd c with trnsversl b); 7 nd (lines d nd c with trnsversl ). nd re corresponding ngles; 3 nd re lternte interior ngles; 5 nd 6 re corresponding ngles 3. nd re corresponding ngles; 3 nd re sme-side interior ngles; 5 nd 6 re lternte interior ngles 5. Skew; nswers m vr. Smple: Since the pths re not coplnr or prllel, the re skew. 7. pirs 9. pirs. Flse; < ED > nd < HG > re skew. 3. Flse; the plnes intersect. 5. Flse; both lines re in plne. 7. m vr. Smple: E illustrtes corresponding ngles, N illustrtes lt. int. ngles. 9. lws 3. sometimes 33. The lines of intersection re. b. Smple: the lines of intersection of wll with the ceiling nd floor (or the lines of intersection of n of the 6 plnes with two different, opposite fces) 35. Yes; 37. D Lesson 3- pp (verticl ngles), 7 (lternte interior ngles), (corresponding ngles) 3. 3 (lternte interior ngles), (corresponding ngles) 5. } b; c } d; (Given) (lternte interior ngles re congruent.) 3 (orresponding ngles re congruent.) 3 (Trnsitive ropert of ongruence) 7. = 0 becuse corresponding ngles re congruent; = 60 becuse sme-side interior ngles re supplementr. 9. = 5, - 50 = 65. 0; 5 = 00, = = 35, = ; ll the ngles re congruent becuse ech pir re verticl ngles, corresponding ngles, or supplementr ngles. 7. } b (Given); m + m = 80 nd m 3 + m = 80 (Smeside interior ngles re supplementr.); (Given); 3 (If two ngles re supplementr to congruent ngles, then the ngles re congruent.) 9. Drwings will vr. onjecture: The trnsversl is perpendiculr to both prllel lines.. m = 8, m = Lesson 3-3 pp < E > } < G > ; onverse of the orresponding ngles Theorem. 3. < > } < HR > ; onverse of the orresponding ngles Theorem. 5. Given b. nd form liner pir. c. ngles tht form liner pir re supplementr. d. 3 e. If corresponding ngles re congruent, then lines re prllel } b; if sme-side interior ngles re supplementr, then the lines re prllel.. } b; if sme-side interior ngles re supplementr, then the lines re prllel. 3. none 5. } b (onverse of the lternte Eterior ngles Theorem) 7. none 9. nd 3 re supplementr (Liner ir ostulte), so 3 7 (ongruent Supplements Theorem). Therefore, / } m (onverse of the orresponding ngles Theorem).. = 0; m = m = =.5; m = m = If lternte eterior ngles re congruent, then the lines re prllel. 5

6 7. m vr. Smples given: j k Given suppl. 9 If lines re, then sme-side int. re suppl. 8 Suppl. Thm. l n If corresp. re, then lines re. 9. L } N; if sme-side interior ngles re supplementr, then the lines re prllel. 3. N } L; if sme-side interior ngles re supplementr, then the lines re prllel Lesson 3- pp suppl. 9 Given. 3 b. Yes; pieces nd re prllel, nd if line is perpendiculr to one of severl prllel lines, it is perpendiculr to ll of the prllel lines. 3. Smple: onsider the edges of rectngulr prism. The top front edge nd the bottom side edge re both perpendiculr to verticl edge of the prism. The top front edge nd the bottom side edge re skew, so the cnnot be prllel. 5. Mesure n three interior ngles to be right ngles nd opposite wlls will be prllel becuse two wlls perpendiculr to the sme wll re prllel. 7. The right tringles must hve cute ngles tht mesure The rungs re prllel to ech other becuse the re ll perpendiculr to the sme side.. The rungs re perpendiculr to both sides. The rungs re perpendiculr to one of two prllel sides, so the re perpendiculr to both sides. This lso mens ll of the rungs re prllel. 3. The rungs re prllel becuse the re ll perpendiculr to one side. 5. Refleive: # ; flse; line does not intersect itself t right ngle. Smmetric: If #b, then b#; true; lines nd b form right ngles. Trnsitive: If #b nd b#c, then #c; flse; if nd c re both perpendiculr to b, then } c or nd c re skew. 7. #d b Theorems 3-8 nd #d b Theorem 3-0. } d b Theorems 3-9 nd H Lesson 3-5 pp. 6. Smple nswer: S U X 3 T Y the rllel ostulte, drw < XY > through T, prllel to SU. ecuse ngles tht form liner pir re supplementr, nd UTX re supplementr. the definition of supplementr ngles, m + m UTX = 80. the ngle ddition ostulte, m UTX = m + m 3. the Substitution ropert, m + m + m 3 = 80. _ U nd 3 _ S becuse if lines re prllel, lternte interior ngles re congruent. ongruent ngles hve equl mesure, so m = m U nd m 3 = m S. So, b the Substitution ropert, m S + m + m U = = = m 3 = 9, m = ; nswers m vr. Smple: 80, 3 = 60, so ech ngle is , 60 3., m vr. Smple: m + m + m = 80 b the Tringle ngle-sum Theorem. It is given tht m = 90, so m + m + 90 = 80. the Subtrction ropert of Equlit, m + m = 90. Thus nd re complementr b the definition of complementr ngles. 7. = 7; m = 35, m = 55, m = = 38; = 36, z = 90; m = 7. Smple nswer: D with right ngle (Given); Drw < DE > through, prllel to (rllel ostulte); D nd E re supplementr (ngles tht form liner pir re supplementr.); m D + m E = 80 (Definition of supplementr ngles); m D = m D + m (ngle ddition ostulte); m D + m + m E = 80 (Substitution ropert); D _ nd E _ (If lines re prllel, then lternte interior ngles re congruent.); m D = m nd m E = m (ongruent ngles hve equl mesure.); m + m + m = 80 (Substitution ropert); m = 90 (Definition of right ngle); m + m + 90 = 80 (Substitution ropert); m + m = 90 (Subtrction ropert of Equlit); nd re complementr (Def. of compl. ngles) E 6

7 3. 0, The bisector is prllel to the side common to the congruent ngles. If the mesure of ech congruent ngle is, then the mesure of the eterior ngle is so the bisector forms ngles of mesure. lternte interior ngles re congruent, so the bisector is prllel to the side common to the congruent interior ngles. 7. H Lesson 3-6 pp onstruct congruent lternte interior ngle, then drw the prllel line. 5. J b b J c c 5. onstructions m vr. Smple using the following segments is given. b 9. b 7. b b 9. onstructions m vr. Smples re given.. 9. D G 3. R S b 7

8 5. 7. c line Yes; if the sum of two numbers is 80 nd one of them is less thn 90, then the other must be greter thn 90. b 9. Not possible; if = b = c, then b = c nd = b + c. So the shorter sides would meet t the midpoint of the longer side, forming segment. 3. D 33. 8; if n is the number of points, then n(n - ) is the number of segments. Lesson 3-7 pp undefined O O. + = -3( - ) = -( + ) or - 3 = -( - ) 5. - = - ( - 6) or - = - ( - ) 7. horizontl: = 7; verticl: = 9. m vr. Smple: = 5, = 6, = 6 5, - 6 = 6 5 ( - 5). (0, 0) 0 0 O (0, 0) 3. 0; the -is is horizontl line, nd the slope of horizontl line is 0; = 0 5. Yes; if the rmp is in. high nd 7 in. long, the slope will be 7 = 0.3 which is less thn the mimum slope of = = ; The slope represents the rte t which the pressure chnges for ech foot of dive. b. ; The -intercept is the vlue of the pressure when the depth of the dive is = 5 b. - 5 = - 5 ( - ) or = c. The bs. vlue of the slopes is the sme, but one slope is pos. nd the other is neg. One -int. is 0 nd the other is No; the slope of the line through the first two points is - 3, nd the slope of the line through the lst two points is -, so the points do not lie on the sme b Lesson 3-8 pp Yes; the slope of / is -, nd the slope of / is -, nd two lines with the sme slope re prllel. 3. = = ( + ) 7. No; the slope of / is -, nd the slope of / is 5. Since the product of the slopes is not -, the lines re not perpendiculr = - 3 ( - 6). - = ( - ) 3. Yes; both slopes re -, so the lines re }. 5. No; the slope of the first line is - 3, nd the slope of the second line is -3. Since the slopes re not equl, the lines re not } No; if two equtions represent lines with the sme slope nd the sme -intercept, the equtions must represent the sme line.. slope of = slope of D = - 3, } D; slope of = slope of D =, } D 3. slope of = slope of D = 0, } D; slope of = 3, slope of D = 3, is not prllel to D Yes; the equtions represent horizontl line nd verticl line, nd ever horizontl line is perpendiculr to ever verticl line. 9. = b. (00, 50) = 3 ( - ) Lesson 3-9 pp neither b. Euclidin c. sphericl 3. ll equingulr tringles in Euclidin geometr hve three 60 ngles. 5. Flse; Drwings m vr. Smple: b. Flse; Drwings m vr. Smple: D D D c. Flse; Drwings m vr. Smple: D 8

9 7. Yes, verticl ngles seem to be congruent in sphericl geometr. Eplntions m vr. Smple: 33. slope: ; -intercept: - O 9. In sphericl geometr, line segment is n rc of gret circle. In Euclidin geometr, line segment is prt of line.. ; Drwings m vr. Smple: 35. = = ( - ) or + = ( - 3) 39. }. } = -6( - 3) 5. sphericl 7. neither 9. sphericl 5. flse 53. The equtor is gret circle. Therefore, it is line in sphericl geometr. TEKS umultive rctice pp. 5. D D H 5. In both Eucliden nd sphericl geometries, tringles hve three sides nd three ngles. In Euclidin geometr, tringles re flt nd the sum of the mesures of their ngles is lws 80. In sphericl geometr, tringles hve curved sides nd the sum of the mesures of their ngles is lws greter thn 80. Topic Review pp remote interior ngles 3. lternte interior ngles 5. 5 nd 8, lines nd b, trnsversl c; nd 6, nd e, trnsversl d 7. nd 7, lines c nd d, trnsversl b 9. lt. int. ngles. m = 75 becuse sme-side int. ngles re suppl; m = 05 becuse lt. int. ngles re n } p; if corresp. ngles re, then the lines re }. 7. / } m; if sme-side int. ngles re suppl., then the lines re }. 9. }. st Street nd 3rd Street re } becuse the re both#to Morris venue. Since st Street nd 5th Street re both } to 3rd Street, st Street nd 5th Street re } to ech other. 3. = 5, = Q 9. b m b 3. - M Q N 9. = 0, ( - 8) = 76, = 35, ( - ) = 69. onverse: If figure hs t lest two right ngles, then it is squre (flse); inverse: If figure is not squre, then it does not hve t lest two right ngles (flse); contrpositive: If figure does not hve t lest two right ngles, then it is not squre (true). 3. It is given tht / } m. b the orresponding ngles Theorem. It is given tht. the Trnsitive ropert of ongruence,. So n } p b the onverse of the orresponding ngles Theorem. Topic Lesson - pp , D, D, D, D, D 3. K KJ 9. O SI, OL ID, LY DE, Y ES. 335 ft ft Yes; two pirs of sides nd two pirs of ngles re mrked s congruent; the third pir of sides re congruent b the Refleive ropert of ongruence, nd the third pir of ngles re congruent b the Third ngles Theorem.. D (Given); it is lso given tht } D, so D becuse the re lternte interior ngles. D b the Third ngles Theorem. D nd D (Given), nd b the Refleive ropert of ongruence. So D b the definition of congruent tringles. 3. m = m D = 0 5. = EF =

10 9. = 5 3. Two pirs of sides re given s congruent, nd the third pir of sides re congruent b the Refleive ropert of ongruence. becuse perpendiculr lines form right ngles, nd ll right ngles re congruent. D D b the lternte Interior ngles Theorem, nd D D b the Third ngles Theorem. So D D b the definition of congruent ngles. 33. Two pirs of congruent sides re given, nd the third pir of sides re congruent becuse Q bisects RT, so TS RS. R } TQ, so Q nd R T becuse the re lternte interior ngles; the third pir of ngles re verticl ngles, so the re congruent. Thus, RS QTS b the definition of congruent tringles. 35. two; (3, ) or (3, 7) 37. It is given tht D nd E, nd so m = m D nd m = m E b the definition of congruent ngles. Thus m + m + m = 80 nd m D + m E + m F = 80 b the Tringle ngle-sum Theorem. the Substitution ropert of Equlit, we hve m + m + m = m D + m E + m F nd m D + m E + m = m D + m E + m F. This gives us m = m E b the Subtrction ropert of Equlit, nd so E, b the definition of congruent ngles Lesson - pp Given b. Refleive ropert of ongruence c. JKM d. LMK 3. Three; nswers will vr. Smple: If ou know the lengths of two sides, ou must know the ngle mesure in between them to form the third side tht would mke the tringles congruent. Knowing the mesurements of two sides or two ngles is not enough since the ngle mesures or side lengths could vr, respectivel. 5. ZD ZD b the Refleive ropert of ongruence, nd it is given tht ZW ZS SD DW. So WZD SDZ b SSS. 7. You need to know RS WU; the digrm shows tht R W nd RT WV. R is included between RT nd RS, nd W is included between WV nd WU. 9. NG RWT b SS.. JEF SFV (or JEF SFV ) b SSS. 3. m vr. Smple: L b. m vr. Smple: L J N M 5. FG } KL (Given), so GFK LKF becuse the re lternte interior ngles. FG KL (Given) nd KF KF (Refleive ropert of ongruence), so FGK KLF b SS D Lesson -3 pp MO NMO b S. 3. VZY VWY b S. 5. Given the perpendiculr segments, Q S becuse perpendiculr lines form right ngles, nd ll right ngles re congruent. It is given tht T is the midpoint of R, so T RT b the definition of midpoint. TQ RTS becuse verticl ngles re congruent, so QT RST b S. 7. Yes; use the Third ngles Theorem. Then ou hve three pirs of congruent ngles nd one pir of congruent sides, so ou cn use the S ostulte. 9. It is given tht N nd MO QO. lso, MON QO becuse verticl ngles re congruent. So MON QO b S.. Given the prllel lines, D nd D becuse lternte interior ngles re congruent. lso, b the Refleive ropert of ongruence. So D b S. 3. E ED, E ED, D D, D 5. It is given tht D nd E. the Third ngles Theorem, F. It is lso given tht DF. It follows tht DEF b S onverse: If ou re too oung to vote in the United Sttes, then ou re less thn 8 ers old; true. K Technolog Lb -3 p. 63 J K. No, ou cnnot use to prove s becuse two s cn hve 3 pirs of s nd not be the sme size. 3. No; the circle intersects > just once, so onl one is formed. If the s re obtuse, then there could be n SS congruenc since cn hve onl one obtuse. M N 0

11 Lesson - pp KLJ OMN b SS; KJ ON, K O, J N. 3. OM ER nd ME RO (Given). OE OE b the Refleive ropert of ongruence. MOE REO b SSS, so M R becuse corresponding prts of congruent tringles re congruent. 5. pir of congruent sides nd pir of congruent ngles re given. Since T T (Refleive ropert of ongruence), ST OT b SS. S O becuse corresponding prts of congruent tringles re congruent. 7. KL bisects KQ, so KL QKL. KL KL b the Refleive ropert of ongruence. KL QKL b SS, so Q becuse corresponding prts of congruent tringles re congruent. 9. LK QLK becuse perpendiculr lines form right ngles, nd ll right ngles re congruent. From the ngle bisector, KL QKL. So with KL KL b the Refleive ropert of ongruence, LK QLK b S nd Q becuse corresponding prts of congruent tringles re congruent.. The construction mkes E, D F, nd D EF. So D EF b SSS. Thus becuse corresponding prts of congruent tringles re congruent. 3. It is given tht JK } Q, so K Q nd J becuse the re lternte interior ngles. With JK Q (Given), KJM QM b S nd then JM M becuse corresponding prts of congruent tringles re congruent. Thus M is the midpoint of J b definition of midpoint. So KQ, which contins point M, bisects J b the definition of segment bisector. 5. Using the given informtion nd E E (Refleive ropert of ongruence), KE E b SSS. Thus KS S becuse corresponding prts of congruent tringles re congruent. In KS nd S, K (Given) nd S S (Refleive ropert of ongruence), so KS S b SS. Thus KS S becuse corresponding prts of congruent tringles re congruent, nd S is the midpoint of K b the definition of midpoint Lesson -5 pp VX ; onverse of Isosceles Tringle Theorem 3. VY ; onverse of Isosceles Tringle Theorem nd Segment ddition ostulte 5. Smple nswer: Yes, the builder is correct. D is given, nd b the definition of ngle bisectors, D. the Refleive ropert of ongruence,. So b the SS ostulte, D. D becuse corresponding prts of congruent tringles re congruent. 7. = 38, = 9. 5, The tringle is n obtuse isosceles tringle; ( + 5) + (3-35) + = 80, so the ngle mesures re 0, 0, nd isosceles tringle b. 900 ft; 00 ft c. The tower is the perpendiculr bisector of ech bse. 9. ( -, 0), (0, 0), (0, - ), (, ), (, 8), (8, ). ( -, 6), (, 6), (, 9), (5, 0), (5, 3), (8, 3) m vr. Smple: You need to know EF to prove the tringles congruent b SS OR ou need to know D to prove the tringles congruent b S OR ou need to know F to prove the tringles congruent b S. Lesson -6 pp If two ngles re congruent nd supplementr, the re right ngles. b. definition of right tringle c. Given d. Refleive ropert of ongruence e. HL 3. E nd DE re right tringles. b. E E c. DE d. HL 5. From the given informtion bout perpendiculr segments, TM nd RMJ re right tringles. M RJ (Given), nd since M is the midpoint of TJ, TM JM. Thus TM RMJ b HL. 7. = -, = 3 9. Yes; two right tringles with congruent hpotenuses nd pir of congruent cute ngles lso hve pir of congruent right ngles. So the two right tringles re congruent b S From the given informtion bout the isosceles tringle, the right ngles, nd the midpoint, ou cn conclude tht KG KE (Definition of isosceles tringle), LKG nd DKE re right tringles (Definition of right tringles), nd LK DK (Definition of midpoint). So LKG DKE b HL, nd LG DE becuse corresponding prts of congruent tringles re congruent. 7. No, the tringles re not congruent. Eplntions m vr. Smple: DF is the hpotenuse of DEF, so it is the longest side of the tringle. Therefore, it is greter thn 5 nd greter thn 3 becuse it is longer thn either of the legs. So DF cnnot be congruent to, which is the hpotenuse of nd hs length No; visulize squeezing t points nd. Tht would chnge the distnce but would not chnge n of the given informtion. Since ou cn chnge the length of, ou cnnot prove to be equilterl.. m vr. Smple: XRY XRZ b HL, YQX YQZ b HL, ZX ZY b HL, nd XS YS b SS

12 Lesson -7 pp M 3. XY 5. divides segment into two congruent segments. So LJM GKM b S. c. If ou use the congruent verticl ngles JML nd KMG, ou cn prove LJM GKM b S. R is common ngle. 7. Given b. Refleive ropert of ongruence c. Given d. S e. orresponding prts of congruent tringles re congruent. 9. QD U nd QD UD (Given), nd D D (Refleive ropert of ongruence). So QD UD b SS.. m vr. Smple: D ED nd D is the midpoint of F (Given). D FD (Definition of midpoint), nd D EDF (Verticl ngles re congruent.). So D EDF b SS, nd E becuse corresponding prts of congruent tringles re congruent. D EDG becuse verticl ngles re congruent, so D EDG b S. 3. The overlpping tringles re E nd D. It is given tht nd. lso, b the Refleive ropert of ongruence. So E D b S. 5. It is given tht E nd D, nd b the Refleive ropert of ongruence. So D E b SS, nd E becuse corresponding prts of congruent tringles re congruent. 7. E D. D, D, E E, DE E b. showing tht D (b S), ou cn then prove tht E ED nd ED E (b S or S). The segments re congruent becuse corresponding prts of congruent tringles re congruent. 9. H. J G M L K b. It is given tht L J } GK, so J K nd L G becuse if lines re prllel, then lternte interior ngles re congruent. lso, LM GM becuse M is the midpoint of LG (Given) nd midpoint Topic Review pp legs 3. corollr 5. ML 7. ST D 7. not enough informtion 9. SS. TVY YWX b S, so TV YW becuse corresp. prts of tringles re. 3. E DE b SSS, so D becuse corresp. prts of _ tringles re. 5. =, = = 65, = LN # KM (Given), so MLN nd KLN re rt. tringles. KL ML (Given) nd LN LN (Refl. rop. of ), so KLN MLN b HL. 3. E D b SS, S, or S 33. TR TS b S TEKS umultive rctice pp Yes, b the onverse of the Isosceles Tringle Theorem. 3. ) T _ SG nd T } SG (Given); ) GT _ TG (Refleive); 3) GT _ TGS (lternte Interior ngles Theorem); ) GT _ TSG (SS) 5. The re congruent right tringles, becuse D #. The shre the leg D, nd hve congruent hpotenuses nd, so the re congruent b HL. 7. For LMNK, LM = KN = nd MN = LK =, while for QRS, Q = SR = 3 nd S = RQ =. The length of rectngle LMNK is units, while the length of rectngle QRS is 3 units, so the rectngles re not. m vr. Smple: To mke the rectngles, ou could chnge the vertices of LMNK so tht M hs coordintes (, 5) nd N hs coordintes (, 3). OR ou could chnge the vertices of QRS so tht R hs coordintes (, -) nd Q hs coordintes (3, -). Topic 5 Lesson 5- pp (, ) 3. (3.5, ) or ( 7, ) 5. ( -.5,.) 7. (0, - ); found 5 of the -distnce to, which is, nd dded it to the -coordinte of ; found 5 of the -distnce from to, which is lso, nd dded it to the -coordinte of b. ( - 3, 0 ) or ( -.5, 0). 5. b. (-, 0.5) mi mi mi Everett, hrleston, rookline, Firfield, Dvenport 3. Yes; Smple: D divides the horizontl distnce between nd into two equl segments, D nd. D lso divides the verticl distnce between nd into two equl segments, DM nd M. D is perpendiculr to D nd to D. Therefore, b SS, DM M.

13 33. m vr. Smple: the formul G( ( - ), ( - ) ) gives G( (8), (-) ) or G(- 5, 3 5 ). Number sense cn help ou dd nd subtrct the frctions given. The coordintes re G(- 5,3 5 ). 35. m vr. Smple: Since ST = ( - ) + ( - ), substituting ( + n) for nd ( + n) for mens ST = ( - ) + (( + n) - ( + n)). nd (( + n) - ( + n)) is simplified s ( - ). So Q = ST. 37. Terminl : (3, 5 ); Terminl : ( -, - 5 ); Terminl : (, 9 ) units. D 3. (-8, 9) b. 7.9 units Technolog Lb 5- pp The length of midsegment of is hlf the length of the third side of the, nd its slope is the sme s the slope of the third side. 3. DE F F, EF D D, DF E E b. tringle s three midsegments divide the into four s. 5. The re of is times the re of ech smll. b. The perimeter of is times the perimeter of ech smll. Lesson 5- pp ft 9 in.; becuse the red segments divide the legs into four congruent prts, the white segment divides ech leg into two congruent prts. The white segment is midsegment of the tringulr fce of the building, so its length is one hlf the length of the bse. 3. 0; ST is midsegment of QR, so b the Tringle Midsegment Theorem, ST } R. Then m QR = m QST becuse corresponding ngles of prllel lines re congruent. 5. Slope of HJ = = nd slope of EF = =. The slopes re equl, so HJ } EF. HJ = ( - ) + (0 - ) = 8 = nd EF = ( - 5) + ( - 6) = 3 =, so HJ = EF G(, ); H(0, ); J(8, 0) 7. The midpoint of FG is J( -6 +, + 8 ) = J(-, 6). The midpoint of FH is K( -6 +, + (-) ) = K(-, ). The slope of JK = - 6 = 5 nd the - - (-) slope of GH = = 5, so JK } GH. JK = (- - (-)) + ( - 6) = + 5 = 6 nd GH = ( - ) + (- - 8) = + 00 = 0 = 6, so JK = GH cm Lesson 5-3 pp. 0. m vr. Smple nswer: onjecture: Qudrilterl D is squre. 3. X M Y Z XMZ YMZ. onjecture: the two right tringles formed b the midpoint of side of n equilterl tringle nd two of its vertices re congruent. 5. M is the perpendiculr bisector of JK No; it is not on the perpendiculr bisector of the street connecting the two schools.. M M (Refleive ropert of ongruence); (Given); M is the midpoint of (Given); M M (Definition of midpoint); M M (SSS ostulte); M M (orresponding prts of congruent tringles re congruent.); m M + m M = 80 (Definition of liner pir); m M = m M = 90 ( M nd M re congruent.); M is the perpendiculr bisector of (Definition of perpendiculr bisector). 3. 7; M is on the perpendiculr bisectors of nd, so M = M nd M = M b the erpendiculr isector Theorem. So M = M = M b the Trnsitive ropert of Equlit. ME ME ME becuse if line is perpendiculr to plne, then it is perpendiculr to ever line in the plne tht contins the intersection of the plne nd the line (nd ll right ngles re congruent). EM EM b the Refleive ropert of ongruence. So EM EM EM b SS. Therefore, E E E, since corresponding prts of congruent tringles re congruent. So E = E = E. 3. Line / through the midpoints of two sides of is equidistnt from,, nd. This is becuse nd 3 b S. D E nd E F becuse corresponding prts of congruent tringles re congruent. the Trnsitive ropert of ongruence, D E F. the definition of congruence, D = E = F, so points,, nd re equidistnt from line /. D E 3 F 3

14 5. the definition of bisector, M M, nd M nd M re right ngles. ecuse ll right ngles re, M M. the Refleive ropert of ongruence, M M. So M M b SS. orresponding prts of congruent tringles re congruent, so nd =. 7. In right SQ nd right SRQ, QS RQS (Given), QS QRS (ll right ngles re congruent.), nd QS QS (Refleive ropert of ongruence). So SQ SRQ b S, nd S SR (or S = SR) becuse corresponding prts of congruent tringles re congruent. 9. H ctivit Lb 5- p. 3. The bisectors of the s of meet t single point. 3. The # bisectors meet t single point, but tht point is outside the. b. m vr. sk our techer to check our work. Lesson 5- pp ( -, - 3) 3. The circumcenter of the tringle formed b the lifegurd chir, the snck br, nd the vollebll court is equidistnt from the vertices of the tringle. lce the reccling brrel t the intersection pt. of two of the tringle s perpendiculr bisectors. S V 5. Z 7. n equilterl tringle 9. m DGE = 30, m DGF = 0, m EGF = 0. From the given informtion, XF XE nd XE XD; lso, XF XE XE XD becuse perpendiculr lines form right ngles, which re congruent. X X nd X X b the Refleive ropert of ongruence, so XF XE nd XE XD b S. Thus XF = XE nd XE = XD becuse corresponding prts of congruent tringles re congruent, nd XF = XD b the Trnsitive ropert of Equlit. Thus X is on the bisector of b the onverse of the ngle isector Theorem. Since n is the bisector of (Given), n contins X. 3. true 5. Method : MK MR nd MKV nd MRV re right ngles. So M is equidistnt from the sides of KVR. It follows tht M is on the ngle bisector. Thus VM is the bisector of KVR. Method : MKV nd MRV re right ngles, so MKV MRV re right tringles. MK MR (Given) nd MV MV (Refl. rop. of ), so MKV MRV (HL). Then KVM RVM since the re corresponding prts of congruent tringles, so VM bisects KVR. Technolog Lb 5-5 p. 9. The pper to be concurrent; es. 3. cute : inside, inside, inside, inside; rt. : on, inside, on, inside; obtuse : outside, inside, outside, inside L Lesson 5-5 pp E 5. DE 7. M L 9. sk our techer to check our work. The folds should show the perpendiculr bisectors of the sides to identif the midpoint of ech side. The should lso go through ech verte nd the midpoint of the opposite side.. 3. TY = 8, TW = 7 5. VY = 6, YX = 3 7. ltitude; it etends from verte of nd is perpendiculr to the opposite side. 9. (6, ). is the intersection of the ltitudes, so it is the orthocenter; is the intersection of the ngle bisectors, so it is the incenter; is the intersection of the medins, so it is the centroid; D is the intersection of the perpendiculr bisectors of the sides, so it is the circumcenter. 3. incenter ircumcenter: outside Incenter: inside entroid: inside Lesson 5-6 pp ssume temporril tht it is not rining outside. 3. ssume temporril tht XY nd re not congruent. 5. I nd II 7. right ngle b. right ngles c. 90 d. 80 e. 90 f. 90 g. 0 h. more thn one right ngle i. t most one right ngle 9. ssume temporril tht / } p. Then = becuse if lines re prllel, then corresponding ngles re congruent. ut this contrdicts the given sttement tht R. Therefore the temporr ssumption is flse, nd we cn conclude tht / is not prllel to p. N

15 . X D ssume temporril tht X X. Since the tringle is sclene, we m suppose tht 6. Find D on T so tht D =. We hve DX X (Given) nd X X (Refleive ropert of ongruence). So DX X b SS. Then DX X (orresponding prts of congruent tringles re congruent.), so DX X (Trnsitive ropert of ongruence). Let m DX = q. Then m X = q (orresponding prts of congruent tringles re congruent.) nd m XD = q (Isosceles Tringle Theorem). Therefore, m X = 80 - q. Now the sum of the ngle mesures in is 80 - q q = 5. This contrdicts the Tringle ngle-sum Theorem. Therefore, the temporr ssumption tht X X is incorrect. So X R X. 3. ssume temporril tht t lest one bse ngle is right ngle. Then both bse ngles must be right ngles, b the Isosceles Tringle Theorem. ut this contrdicts the fct tht tringle is formed, becuse in plne two lines perpendiculr to the sme line re prllel. Therefore the temporr ssumption is flse tht t lest one bse ngle is right ngle, nd we cn conclude tht neither bse ngle is right ngle. 5. ssume tht lines / nd m intersect t point. Given the ngle mesures in the digrm, the mesure of QR is 80 - (mesures of ngles in liner pir re supplementr). Then the sum of mesures in RQ is m RQ, which is greter thn 80 since m RQ must be greter thn zero. This contrdicts the Tringle ngle-sum Theorem, so therefore lines / nd m don t intersect nd the re prllel. 7. D 9. ll cute tringles; the centroid nd incenter re lws inside. The circumcenter nd orthocenter re on the tringle if it is rt. tringle; the re outside the tringle if it is obtuse. Lesson 5-7 pp XZ + ZY 7 XY X X Z Y Z Y 3. Smple nswer: You cn verif tht the Tringle Inequlit Theorem works for prticulr tringle, while the proof of the Tringle Inequlit Theorem works for ll given tringles. 5. This is true b the orollr to the Tringle Eterior ngle Theorem. 7. (, ), (, 5), (, 6), (3, 3), (3, ), (3, 5), (3, 6), (3, 7), (, 3), (, ), (, 5), (, 6), (, 7), (, 8) 9. The dshed red line nd the courtrd wlkw determine three sides of tringle, so b the Tringle Inequlit Theorem, the pth tht follows the dshed red line is longer thn the courtrd wlkw.. No; + 3 is not greter thn Yes; , , nd m 6 6 m 7. lce the computer t the corner tht forms right ngle; plce the bookshelf long the wll opposite the right ngle. In right tringle the right ngle is the lrgest ngle, nd the longest side of tringle is opposite the lrgest ngle. 9. G, H, J. TU, UV, TV 3. YZ, XZ, XY 5. m vr. Smple: The sum of the interior ngle mesures of tringle is 80, so m T + m + m = 80. Since m T = 90, m + m = 90, nd so m T 7 m (omprison rop. of Inequlit). Therefore 7 T b Theorem Lesson 5-8 pp D 3. onverse of the Isosceles Tringle Theorem b. Given c. Def. of midpt. d. = D e. Given f. Hinge Theorem E D (Given) so E nd D re isosc. with = E = D =. Since m ED 7 m E (Given), ED 7 E b the Hinge Thm. 9. In O nd O, O = O = 7, O = O = 5, = 8, nd = 68. Since 7, m O 7 m O b the onverse of the Hinge Thm.. J 3. D Topic Review pp medin 3. incenter (0, 0) 9. (6, -). (-6, -7) Let point S be second bse nd point T be third bse. Find the midpt. M of ST nd then through M construct the line / # to ST. oints of the bsebll field tht re on line / re equidistnt from second nd third bse (3, ) 5. (5, ) is n ltitude; it is segment from verte tht is # to the opposite side. 3. QZ = 8, QM = 33. (, -3) 35. ssume temporril tht the third line intersects neither of the first two. Then it is } to both of them. Since the first two lines re } to the sme line, the re } to ech other. This contrdicts the given informtion. Therefore the temporr ssumption is flse, nd the third line must intersect t lest one of the two others. 37. ssume temporril tht n equilterl tringle hs n obtuse ngle. Since ll the ngles re in n equilterl tringle, then ll three ngles must be obtuse. ut we showed in E. 36 tht tringle cn hve t most one obtuse ngle. Therefore the temporr ssumption is flse, nd n equilterl tringle cnnot hve n obtuse ngle. 39. RS, ST, RT. Yes; , , nd

16 TEKS umultive rctice pp / nd m do not hve n points in common. b. m vr. Smple: Use the Lw of Detchment to conclude tht / nd m do not intersect. Then use the Lw of Detchment gin to conclude tht / nd m do not hve n points in common. 5. ssume temporril tht is rt. tringle, s well s obtuse (Given). Then one ngle mesure is 90 (Def. of rt. tringle), nd one ngle mesure is 7 90 (Def. of obtuse tringle). The sum of these two mesures is This contrdicts the Tringle ngle-sum Thm., so the temporr ssumption is flse. Hence, is not rt. tringle. Topic 6 Technolog Lb 6- p. 8. m vr. Smple: s form circle, which hs 360 degrees. 3. sk our techer to check our work. b. regulr, regulr hegon c. 3 sides: 0; sides: 90; 5 sides: 7; 6 sides: 60; 8 sides: 5 d. m vr. Smple: To tile plne, the mesure of et. must be fctor of 360. Lesson 6- pp , 80, 80; 80, 50, w = 7, = 59, = 9, z = 9. 80n ; (n - ) # 80 b. 80n - [(n - ) # 80] = 360 ; olgon Et. ngle-sum Theorem. Yes; nswers m vr. Smple: The sum of the mesures of interior ngles = (n - ) # , 5, You should select geometr softwre, becuse the cn use it to mke regulr polgons nd mesure ngles. b. onjecture : The sum of the mesures of the eterior ngles of regulr polgon, one t ech verte, is 360. onjecture : The mesures of the eterior ngles of regulr polgon re equl Lesson 6- pp = 5, = definition of prllelogrm b. If lines re prllel, then lternte interior ngles re congruent. c. Opposite sides of prllelogrm re congruent. d. E DE e. orresponding prts of congruent tringles re congruent. f. nd D bisect ech other t E. 3. = 5, = 7 5. m vr. Smple: Suppose RSTW nd XYTZ re prllelogrms. It follows tht R T nd T X, since opp. ngles of re. the Trnsitive rop. of, R X. 7. m = 38, m = 3, m 3 = 0 9. m = 95, m = 37, m 3 = 37. Since D is prllelogrm, it follows from the def. of tht } D nd } D. Since sme-side int. ngles re suppl. is suppl. to nd is suppl. to D. 3. m vr. Smple: It is given tht } D nd D } EF. constructing G } nd DH } E, it follows from the Def. of tht G nd DHE re prllelogrms. G nd E DH since opp. sides of re. It is given tht E, so b the Trns. rop. of, G DH. lso, G } DH becuse if two lines re } to the sme line, then the re } to ech other. lso, 3 6 nd GD HDF, nd if lines re }, corresp. ngles re. So, b S, GD HDF. Finll, D DF since corresp. prts of tringles re. 5. D 7. Lesson 6-3 pp m vr. Smple:. D, DE F, E F (Given). DE = F, E = F (Def. of ) 3. DE + E = F + F (dd. rop. of = ). D = (Segment ddition ostulte) 5. D (Def. of ) 6. D is. (If both pirs of opposite. sides of qud. re, then the qud. is.) 3. m vr. Smple:. M is the midpoint of HK nd JL. (Given). HM KM, JM LM (Def. of midpoint) 3. HK nd JL bisect ech other. (Def. of bisect). HJKL is. (If the digonls of qud. bisect ech other, then the qud. is.) 5. =, = = 3, =. m vr. Smple: E F D H G 3. No; onl one digonl is bisected. 5. The connecting pieces D nd re congruent, nd the distnces nd D between where the two pieces ttch re equl. The side lengths of D do not chnge s the tckle bo opens nd closes. Since both pirs of opposite sides re congruent, D is lws prllelogrm. the definition of prllelogrm, is prllel to D, so the shelves re lws prllel to ech other. 7. m vr. Smple:. is suppl. to. (Given). } D (If sme-side int. ngles re supplementr, then lines re }.) 3. is suppl. to D. (Given). } D (If sme-side int. ngles re supplementr, then lines re }.) 5. D is. (Def. of ) 9. F. m vr. Smple:. NRJ T (Given). NJ T (orresp. prts of s re.) 3. JN } T (Given). JNT is. (If one pir of opp. sides of qud. is both nd }, then the qud. is.) Lesson 6- pp m = 55, m = 35, m 3 = 55, m = m = 90, m = 55, m 3 = = 3; LN = M = 7 7. = 9; LN = M = =.5; LN = M =.5. Rectngle; the prllelogrm hs right ngles nd does not hve congruent sides. 3. given b. def. of rectngle c. Refl. rop. of d. def. of rectngle e. D f. D g. ll rt. ngles re. h. orresp. 6

17 prts of tringles re. 5. m H = m J = 58, m K = m G =, HK = KJ = JG = GH = 6 7. = D = 6 9. = D =. = 3, = 5; ll sides re m vr. Smple:. LN is rectngle. (Given). LN (Digonls of rectngle re.) 3. T = TL = TN = T (Digonls of bisect ech other.). LT TN (Vert. ngles re.) 5. LT TN (SS) Lesson 6-5 pp E E nd DE E; No. The digonls bisect ech other, so b Theorem 6-, the qudrilterl is prllelogrm. However, ou do not know if the digonls re congruent, so ou do not hve enough informtion to show tht the qudrilterl is rectngle. b. D, D, nd m D = 90; Yes. Theorem 6-8, since opposite sides re congruent, the qudrilterl is prllelogrm. ecuse m D = 90, the qudrilterl is rectngle. c. } D, D }, nd D; Yes. The qudrilterl is rectngle b Theorem 6-8. d. E E DE E; Yes. The digonls bisect ech other, so b Theorem 6-, the qudrilterl is prllelogrm. The digonls re congruent, so b Theorem 6-8, the qudrilterl is rectngle. 9. Squre; since it is rhombus nd rectngle, it must be squre.. Rhombus; digonls re perpendiculr. 3. Rectngle, squre; nswers m vr. Smple: 5. Rectngle, rhombus, squre; nswers m vr. Smple: b b b b b b 7. m vr. Smple:. D, D (Given). D (Opposite sides of re.) 3. D D (Refl. rop. of ). D D (SSS) 5. D D (orresp. prts of tringles re.) 6. m D + m D = 80 (Sme-side int. ngles re suppl.) 7. D nd D re rt. ngles. (Def. of rt. ngle) 8. D is rectngle. (Def. of rectngle) 9. onstruct the midpt. of digonls. op the digonls so the two midpts. coincide. onnect the endpoints of the digonls.. No; if the digonls of re, then it would hve to be rectngle nd hve rt. ngles. 3. No; in, consecutive ngles must be suppl., so ll ngles must be rt. This would mke it rectngle Lesson 6-6 pp m vr. Smple:. Drw E } D. (onstruction). ED is prllelogrm. (Def. of prllelogrm) 3. E D (Opp. sides of prllelogrm re congruent.). (If lines re prllel, corresp. ngles re congruent.) 5. (Isosc. Tringle Thm.) 6. (Trnsitive rop. of ongruence) 7. D nd re suppl. (Sme-Side Int. ng. ost.) 8. D nd re suppl. (Sme-Side Int. ng. ost.) 9. D D (ngles suppl. to congruent ngles re congruent.) No; eplntions m vr. Smple: ssume KM bisects both ngles. Then MKL MKN KML KMN. oth pirs of sides of KLMN would be prllel, nd KLMN would be prllelogrm. It is impossible for n isosceles trpezoid to lso be prllelogrm, so KM cnnot bisect LMN nd LKN. 7. m = m = 60, m D = 0 9. m vr. Smple: Given: Trpezoid D with } D, D rove: D is n isosc. trpezoid. There re two cses to consider, whether nd D re cute or obtuse. nd D cnnot be right ngles; if the were, then D would be rectngle, not trpezoid. se I: nd D re cute.. < > is not } to < D >. (Def. of trpezoid). Etend < > nd < D > to meet t point T. (onstruction) 3. D (Given). T DT (onverse of Isosc. Thm.) 5. T nd T D. (If } lines, then corresp. ngles re.) 6. T T (Trnsitive rop. of ) 7. T T (onverse of Isosc. Thm.) 8. T + = T, T + D = TD (Seg. dd. ost.) 9. = T - T, D = TD - T (Subtr. rop. of Eq.) 0. = D (Subst. rop. of Eq.). D is n isosc. trpezoid. (Def. of isosc. trpezoid) se II: nd D re obtuse.. is supplementr to nd is supplementr to D (Sme-Side Int. ngles ost.). (ongruent Suppl. Thm.) 3. nd re cute (suppl. to obtuse ngles). D is n isosc. trpezoid. (onverse of Thm. 6 0, se I) 7

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