Answers to Exercises ABC ABC

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1 nswers to ercises HTR HTR LSSON. HTR HTR. postulte is sttement ccepted s true without proof. theorem is deduced from other theorems or postultes.. Sutrction: quls minus equls re equl. Multipliction: quls times equls re equl. ivision: quls divided y nonzero equls re equl.. Refleive: ny figure is congruent to itself. Trnsitive: If Figure is congruent to Figure nd Figure is congruent to Figure, then Figure is congruent to Figure. If, then. Q R S X Y Q X Y If Q RS nd RS XY, then Q XY. Symmetric: If Figure is congruent to Figure, then Figure is congruent to Figure. If then Y M X L Z N M Y L X N Z,. If XYZ LMN, then LMN XYZ. 4. refleive property of equlity; refleive property of congruence 5. trnsitive property of congruence 6. sutrction property of equlity 7. division property of equlity 8. istriutive; Sutrction; ddition; ivision 9. Given; ddition property of equlity; Multipliction property of equlity; ommuttive property of ddition. 0. true, definition of midpoint. true, Midpoint. true, definition of ngle isector. true, ngle isector 4. flse, Line Intersection 5. flse, Line 6. true, ngle ddition 7. true, Segment ddition 8. Tht ll men re creted equl. Tht they re endowed y their cretor with certin inlienle rights, tht mong these re life, lierty, nd the pursuit of hppiness. Tht to secure these rights, governments re instituted mong men, deriving their just powers from the consent of the governed. Tht whenever ny form of government ecomes destructive to these ends, it is the right of the people to lter or to olish it, nd to institute new government, lying its foundtion on such principles nd orgnizing its powers in such form s to them shll seem most likely to effect their sfety nd hppiness. 9. See flowchrt elow. 0. See flowchrt elow. nswers to ercises 9. (Lesson.) O nd O re rdii Given efinition of circle 0. (Lesson.) O O O is isosceles efinition of isosceles m n Given orresponding ngles 4 orresponding ngles NSWRS TO XRISS 4

2 . See flowchrt elow.. See flowchrt elow.. (n ) (m ) n m (n m ) 4. (n )(m ) 4nm n m 4nm n m (nm n m ) 5. Let n e ny integer. Then the net two consecutive integers re n nd n. The sum of these three integers is (n) (n ) (n ) n n n. omining like terms: n (n ), which is divisile y m 7. sphere, cylinder, cone; cylinder, cone, sphere; cone, cylinder, sphere 8. 0 ft 9. FG 6 nd G ecuse GF nd G re prllelogrms. Tringle FG is right (mfg 90 ) y the onverse of the ythgoren ecuse 6 6.ut mfg 8 nd mg 40 y conjectures regrding ngles in prllelogrm. So, mfg 9 ecuse the sum of the ngles round G is 60. So, mfg is oth 90 nd , y 6, 7. m c. nswers to ercises. (Lesson.) 4 Given Given SSS ongruence 5 T Refleive property of congruence. (Lesson.) onstruct ngle isector ngle isector Given 5 4 efinition of ngle isector Refleive property of congruence SS ongruence 6 T 44 NSWRS TO XRISS

3 LSSON.. Liner ir. rllel, ngle ddition, Liner ir, orresponding ngles. rllel 4. erpendiculr 5. Use the definition of supplementry ngles nd the sustitution property to get m m 80. Then use the division property nd the definition of right ngle Use the definition of supplementry ngles nd the trnsitive property to get m m m m4. Then use the sustitution property nd the sutrction property to get m m4. 7. Use the definition of right ngle nd the trnsitive property to get m m. Then use the definition of congruence to get. 8. Use the V nd the trnsitive property to get. Therefore the lines re prllel y the Use the V nd the trnsitive property to get 4. Therefore the lines re prllel y the onverse of the I.. Use the Liner ir nd the definition of supplementry ngles to get m m 80. Then use the nd the sustitution property to get m m 80.. Use the Liner ir nd the definition of supplementry ngles to get m m m m. Then use the sutrction property nd the onverse of the I to get.. 4 onstruct trnsversl cross lines nd ;it will intersect line y the rllel. Use the Interior Supplements nd the definition of supplementry ngles to get m m 80. Use the nd the sustitution property to get m m 80. Therefore y the onverse of the Interior Supplements. nswers to ercises Use the V nd the to get nd. Then use the trnsitive property to get. NSWRS TO XRISS 45

4 nswers to ercises 4. Use the definition of perpendiculr lines nd the trnsitive property to get m m. Therefore lines nd re prllel y the onverse of the I. 5. y the Tringle Sum, m m m 80. y the definition of right tringle, is right ngle. y the definition of right ngle, m 90. Using the sutrction property, m m 90, so y the definition of complementry ngles, nd re complementry. 6. Liner ir ost. V Thm. onverse of I Thm. onverse of Thm cm ost. 8. re (ft ) ottom 6 sides 9 ck nd front 6 rooftops 8 gle ends Totl lywood (4 y 8) The re is less thn tht of one sheet of plywood. However, it is impossile to cut the correct size pieces from one piece. This nswer ssumes tht the ottom of the dog house is included. If the ottom is not included, nd the gles cn e cut seprtely from the rectngulr prt of the front nd ck, then the pieces cn e cut from single sheet of plywood. 9. (6, 0), (, 6), (0, 0); mpping rule: (, y) ( 8, y ) 46 NSWRS TO XRISS

5 LSSON.. se : is colliner with nd. Use the Line Intersection to show tht nd re the sme point nd use the definitions of isector nd midpoint to get. se : is not colliner with nd. Use the SS ongruence to get. Then use T to get.. se : is colliner with nd. Use the definitions of congruence nd midpoint to show tht is the midpoint of.then use the definition of perpendiculr isector. se : is not colliner with nd.rw midpoint nd. Use the SSS ongruence to get. Then use T nd the ongruent nd Supplementry to prove tht nd re oth right ngles. Therefore is the perpendiculr isector of y the definitions of midpoint nd perpendiculr.. Use the refleive property nd the SSS ongruence to get. Therefore, y T. 4. Use the refleive property nd the S ongruence to get. Then use T nd the definition of isosceles tringle. 5. rw line. Use the Isosceles Tringle to get. Use the ngle ddition nd the sutrction property to get. Then use the onverse of the Isosceles Tringle ( ), the refleive property, nd the SSS ongruence to get. Therefore, y T. 6. m Use the Line Intersection nd the erpendiculr isector to get nd. Then use the trnsitive property nd the onverse of the erpendiculr isector to prove tht point is on line n. 7. Use the Line Intersection nd the ngle isector to prove tht Q is eqully distnt from nd nd from nd. Then use the trnsitive property nd the onverse of the ngle isector to prove tht point Q is on line n. 8. Use the Liner ir nd the definition of supplementry ngles to get m m4 80. Then use the Tringle Sum nd the trnsitive property to get m m m m m4. Therefore, m m m4 y the sutrction property. 9. n m 4 Q n 4 Use the Tringle Sum nd the ddition property to get m m m m m4 m 60. Then use the ngle ddition nd the sustitution property to get m m m m Use the definitions of medin nd midpoint to get M nd N N M. Then use the multipliction property nd the sustitution property to get N M. y the refleive property, the Isosceles Tringle, nd the SS ongruence, N M. Therefore, N M y T. NSWRS TO XRISS 47 nswers to ercises

6 . Q medin y T nd the definitions of midpoint nd medin. nswers to ercises Use the ngle ddition nd the definition of ngle isector to get m m nd mq m. Then use the Isosceles Tringle, the multipliction property, nd the sustitution property to get Q. y the refleive property nd the S ongruence, Q. Therefore, Q y T.. Use the Isosceles Tringle, the Right ngles re ongruent, nd the S to get T S. Therefore, S T y T.. medin ngle isector Use the definitions of medin, midpoint, nd isosceles tringle, the refleive property, nd the SSS ongruence to prove tht. Then use T nd the definition of ngle isector. T S ngle isector ltitude Use the definitions of ngle isector nd isosceles tringle, the refleive property, nd the SS ongruence to get. Then use T, the Liner ir, nd the ongruent nd Supplementry to prove tht nd re oth right ngles. Therefore is the ltitude y the definitions of perpendiculr nd ltitude. 4. 6, y 5..9 m 6. first imge: (0, ); second imge: (6, ) 7. F mkes F rhomus, so its digonls re perpendiculr. mfg 90, so mfg mfg 90. F G, so y sutrction nd trnsitivity m mfg. FG y I, so c One possile sequence:. Fold onto nd crese. Lel s. Lel the midpoint of the rc M.. Fold line onto itself so tht M is on the crese. Lel s. M is the midpoint of, nd is the desired tngent. M ltitude medin Use the Right ngles re ongruent, the Isosceles Tringle, nd the S to get. Therefore, is the 0. m... ( c), y ( ), z ( c). w ( ), ( e), y (e d), z (d ) 48 NSWRS TO XRISS

7 LSSON Use to represent the mesures of one pir of congruent ngles nd y for the other pir. Use the Qudrilterl Sum nd the division property to get y 80. Therefore, the opposite sides re prllel y the onverse of the Interior Supplements.. Use the I, the refleive property, nd the SS ongruence to get. Then use T nd the onverse of the I to get.. Use the definition of rhomus, the refleive property, nd the SSS ongruence to get. Then use T nd the definition of ngle isector to prove tht isects nd. Repet the steps ove using digonl. 4. Use the definition of prllelogrm to get nd. Then use the Interior Supplements Use the refleive property nd the SSS ongruence to get. Then use T nd the onverse of the I to get nd. Therefore, is rhomus y the definitions of prllelogrm nd rhomus. Use the onverse of the Opposite ngles to prove tht is prllelogrm. Then use the definition of rectngle. 7. Use the definition of rectngle to prove tht is prllelogrm nd. Then use the rllelogrm Opposite Sides, the refleive property, nd the SS ongruence to get. Finish with T. 8. Use the rllelogrm Opposite Sides, the refleive property, nd the SSS ongruence to get. Repet the ove steps to get nd. Then use T nd the trnsitive property to get. Finish with the Four ongruent ngles Rectngle. 9. Use the rllel to construct. Then use the rllelogrm Opposite Sides nd the trnsitive property to prove tht is isosceles. Therefore, y the Isosceles Tringle, the, nd sustitution. 0. Use the Isosceles Trpezoid, the refleive property, nd the SS ongruence to get. Then y T. nswers to ercises NSWRS TO XRISS 49

8 S ongruence Line ngle ddition nswers to ercises Use the rllelogrm Opposite ngles, the multipliction property, nd the definition of ngle isector to get. Then use the onverse of the Isosceles Tringle, the definition of isosceles tringle, nd the rllelogrm Opposite Sides to get.. X W Use the onverse of the ngle isector to prove tht WY is the ngle isector of Y. In like mnner, WY is the ngle isector of W. Therefore, WXYZ is rhomus y the onverse of the Rhomus ngles.. Liner ir Y Q Interior Supplements rllelogrm onsec. ngle Z rllelogrm igonl Lemm Opposite Sides onverse of the IT Opposite ngles onverse of the Rhomus ngles oule-dged Strightedge SSS ongruence IT onverse of the ngle isector S 5c. S 5d. N ft 7. See tle elow. 8. V V hs length.8 nd ering c. 5 0d. 0e (Lesson.4) prllelogrm digonls isectors of sides isosceles trpezoid Nme Lines of symmetry Rottionl symmetry none -fold trpezoid none none kite digonl none squre 4-fold rectngle isectors of sides -fold rhomus digonls -fold isector of sides none 50 NSWRS TO XRISS

9 LSSON.5. : ris is in Frnce; Tucson is in the U.S.; London is in nglnd. mko must e the cpitl of Mli.. : The Sir in prt shows tht Hlley ws nglish; Julius esr ws n emperor, not scientist; Mdonn is singer. Glileo Glilei must e the nswer.. No, the proof is climing only tht if two prticulr ngles re not congruent, then the two prticulr sides opposite them re not congruent. It still might e the cse tht different pir of ngles re congruent nd tht therefore different pir of sides re congruent. 4. Yes, this sttement is the contrpositive of the conjecture proved in mple, so they re logiclly equivlent. 5.. ssume the opposite of the conclusion;. Tringle Sum ;. Sustitution property of equlity; 4. 0 ; Sutrction property of equlity 6. ssume ZOI is equingulr. Use the definition of equingulr nd the Four ongruent ngles Rectngle to prove tht ZOI is rectngle. Therefore ZOI is prllelogrm, which cretes contrdiction. 7. ssume is the ltitude to. Use the definitions of ltitude, medin, nd midpoint, the Right ngles re ongruent, nd the SS ongruence to get. Therefore, which cretes contrdiction. 8. ssume ZO I. Use the Opposite Sides rllel nd ongruent to prove tht ZOI is prllelogrm, which cretes contrdiction. 9. Given: ircle O with chord nd perpendiculr isector Show: psses through O ssume does not pss through O. Use the Line to construct O nd O nd the erpendiculr to construct O. Then use the Isosceles Tringle, the Right ngles re ongruent, nd the S to get O O. From T nd the definition of midpoint, prove tht is the midpoint of, which cretes contrdiction , 47, c ft ft N c. S d. S e. O nswers to ercises NSWRS TO XRISS 5

10 . LSSON.6 se The sme: Use the Inscried ngle nd the trnsitive property to get. Z W X Use the yclic Qudrilterl, the Opposite ngles, nd the ongruent nd Supplementry to get R,,, nd T re right ngles. 5. S O T nswers to ercises se ongruent: Use the multipliction property to get myz mwx. Then follow the steps in se.. Use the Inscried ngle, the ddition property, nd the distriutive property to get m m m m.then use the definition of degrees in circle, the sustitution property, nd the definition of supplementry ngles to get nd re supplementry. Repet the steps using m nd m to get nd re supplementry.. Use the Line to construct. Then use the I, the Inscried ngle, nd sustitution to get. 4. T W X Y Use the Line to construct OS,OT, nd O. Then use the Tngent, the onverse of the ngle isector, nd the S to get OS OT. S T y T. 6. Use the Line to construct. Then use the Inscried ngle, the ddition property, nd the distriutive property to get m m m m. Therefore, m m m y the Tringle terior ngle nd the trnsitive property. 7. O Intersecting Secnts : The mesure of n ngle formed y two secnts intersecting outside circle is hlf the difference of the mesure of the lrger intercepted rc nd the mesure of the smller intercepted rc. Use the Tringle terior ngle nd the sutrction property to get. Then use the Inscried ngle, the sustitution property, nd the distriutive property to get m m. R 5 NSWRS TO XRISS

11 8. Use the Inscried ngle to get m m.y the definition of semicircle, m m 80, so y the division property, m 90. y the definition of right ngle, is right ngle. ecuse only definitions, properties, nd the Inscried ngle re needed to prove this conjecture, it is corollry of the Inscried ngle Line SSS ongruence IT Liner ir onverse of the IT S ongruence onverse of the ngle isector terior ngle Sum Inscried ngle yclic Qudrilterl S V ngle ddition Tngent Segments rllel I Tringle Sum SSS ongruence Right ngles re ongruent Tngent rllelogrm igonl Lemm Opposite ngles IT rllelogrm Inscried in ircle ngle ddition S ongruence SS ongruence Midpoint erpendiculr Right ngles re ongruent rc ddition ongruent nd Supplementry. (4.0,.9), (.,.8). ; ; 9.,,,, s long s point is inside the tringle, c h. roof: Let e the length of side. The res of the three smll tringles re,, nd c. The re of the lrge tringle is h. So, c h. ivide oth sides y. So, c h. h c c. 67 6d. 6e. cnnot e determined 6f. cnnot e determined 7. O M Steps:. onstruct O.. isect O. Lel midpoint M.. onstruct circle with center M nd rdius M. 4. Lel the intersection of the two circles nd. 5. onstruct nd. nd re the required tngents. O nd O re right ngles ecuse they re inscried in semicircles. nswers to ercises NSWRS TO XRISS 5

12 LSSON T V nswers to ercises Use the Right ngles re ongruent, ST, nd the Similrity to get T HT GV. Therefore, G V TH V y SST. See the Solutions Mnul for the fmily tree.. G S Use the definitions of medin nd midpoint, the Segment ddition, the sustitution property, nd the multipliction property to get Y I nd SL SM. Then use ST, SST, nd the SS Similrity to get YG SL. Therefore, G S G Y L y SST. See the Solutions Mnul for the fmily tree.. F Use the definition of ngle isector, the ngle ddition, nd the sustitution property to get m m nd mf m. Then use ST nd the Similrity to get QF. Therefore, F F Q y SST. 4. Use the nd the Similrity to get. Then use SST nd the Segment ddition to get. Therefore, y lger. 5. H L M 4 Use the ddition property to get. Then use lger nd the Segment ddition to get. Therefore y the SS Similrity, y ST, nd y the. G Q Y I L Use the Right ngles re ongruent, the refleive property, nd the Similrity to get nd. Therefore, y the trnsitive property of similrity. 7. Use the Three Similr Right Tringles to get. Then use SST to get h h y. 8. rw the ltitude to the hypotenuse, then use the rtios given y the Three Similr Right Tringles c. In prticulr, c d yields c cd. Now look t the other smll tringle nd use d c to get cd. 9. egin y constructing second tringle, right tringle F (with right ngle), with legs of lengths nd nd hypotenuse of length. The pln is to show tht c, so tht the tringles re congruent. Then show tht nd re congruent. Once you show tht is right ngle, then is right tringle. 0. H L Y d h c c y c d Use the ythgoren to write epressions for the lengths of the unknown legs. Show tht the epressions re equivlent. The tringles re congruent y SSS or SS. G F 54 NSWRS TO XRISS

13 ... erpendiculr Segment upliction rllel Similrity SS ongruence SS Similrity SSS Similrity Three Similr Right Tringles ythgoren Segment ddition rllel/roportionlity Similrity Right ngles re ongruent SSS ongruence Similrity 4. pproimtely.5 cm or 6.5 cm c. 5d. 5e. 6. The vectors re digonls of your qudrilterl rottion out the midpoint of the common side; the entire tesselltion mps onto itself rottion out the midpoint of ny side 7. possile nswer: vector running from ech verte of the qudrilterl to the opposite verte (or ny multiple of tht vector) c. 57 8d. 6 8e. SQ nd TQ re 90 y the Tngent nd so re supplementry. So, ST nd SQT must lso e supplementry y the Qudrilterl Sum. Therefore, opposite ngles re supplementry, so it s cyclic. 8f. ecuse S is tngent, msq 90 (Tngent ). ecuse msq 90, SQ must e tngent (onverse of the Tngent ) G FG y S; G G y SS; G FG y the HL 0. F nd F y T; F F (ddition property of equlity). Therefore,, nd is isosceles. 0c. The figure is inccurte. 0d. The ngle isector does not intersect the perpendiculr isector inside the tringle s shown, ecept in the specil cse of n isosceles tringle, when they coincide.. 7 cm, 46 cm, 0 stones. rw the trpezoid nd etend the legs until they meet to form tringle. Use prllel proportionlity to find the rise. The spn is twice the rise. Use inverse tngent to find the centrl ngle mesure. ivide into 80 to find the numer of voussoirs. nswers to ercises NSWRS TO XRISS 55

14 nswers to ercises USING YOUR LGR SKILLS. (, 0). (, c).,,, 4. possile nswer: T (0, ) slope of R slope of 0 (undefined) slope of T slope of TR (undefined) Opposite sides hve the sme slope nd re therefore prllel y the prllel slope property. Two sides re horizontl nd two sides re verticl, so the ngles re ll congruent right ngles. RT is n equingulr prllelogrm nd is rectngle y definition. 5. possile nswer: T (0, 0) y y R (0, 0) X (, 0 ) I (, c) (, ) (, 0) Z (, c ) Y ( +, c ) R (, 0) Let X e the midpoint of TR. X 0, 0 0,0 Let Y e the midpoint of RI. Y, 0 c c, Let Z e the midpoint of TI. Z 0, c 0, c X, Y, nd Z re the midpoints of TR,RI nd TI, respectively, y the coordinte midpoint property. So XY,YZ, nd ZX re midsegments y definition. 6. possile nswer: y (c, d) T (0, 0) slope of TR 0 0 d 0 d slope of R c c d d 0 slope of c c 0 slope of T d 0 c 0 d c TR nd hve the sme slope nd re prllel y the prllel slope property. So TR hs only one pir of prllel sides nd is trpezoid y definition. T (c ) 0 (d 0) c d R ( c ) (d 0) (c) d c d The nonprllel sides of the trpezoid hve the sme length. So trpezoid TR is isosceles y definition. 7. y (, 0) ( c, d) R (, 0) U (0, ) Q (, 0) Show: QU is equilterl Q U ( 0) ) (0 4 UQ ( ) Q U UQ QU is equilterl 56 NSWRS TO XRISS

15 8. Tsk : Given: rectngle with oth digonls Show: The digonls re congruent Tsk : y T (0, ) Tsk : Given: Rectngle RT with digonls R nd T Show: R T Tsk 4: To show tht two segments re congruent, you use the distnce formul to show tht they hve the sme length. Tsk 5: R ( ) 0 ( 0) T ( ) 0 (0 ) So R T ecuse oth segments hve the sme length. Therefore the digonls of rectngle re congruent. 9. Tsk : Given: tringle with one midsegment Show: The midsegment is prllel to nd hlf the length of the third side Tsk : y T (0, 0) R (0, 0) Z ( c, ) d I (c, d) (, ) (, 0) Y ( + c, ) R (, 0) Tsk : Given: TRI nd midsegment YZ Show: YZ TR nd YZ TR Tsk 4: To show tht two segments re prllel, use the prllel slope property. The segments re horizontl, so to compre lengths, sutrct their -coordintes. Tsk 5: Slope of TR 0 0 d Slope of YZ d 0 0 c c The slopes re the sme, so the segments re prllel y the prllel slope property. TR 0 YZ c c TR d So the midsegment is hlf the length of the third side. Therefore the midsegment of tringle is prllel to the third side nd hlf the length of the third side. 0. Tsk : Given: trpezoid Show: The midsegment is prllel to the ses Tsk : y One possile set of the coordintes for TR is shown in the figure. y the coordinte midpoint property, the coordintes of M re, c nd of N re d c,. Tsk : Given: Trpezoid TR with midsegment MN Show: MN TR Tsk 4: To show tht the midsegment nd ses re prllel, you need to find their slopes. Tsk 5: slope of MN c c 0 0 d d slope of TR 0 0 c c 0 The slope of is d d 0. The slopes re equl, therefore the lines re prllel.. Tsk : Given: qudrilterl in which only one digonl is the perpendiculr isector of the other Show: The qudrilterl is kite Tsk : y T (, 0) (, c) (d, c) M T (0, 0) N I (0, ) M(0, 0) (0, c) R (, 0) K(, 0) Tsk : Given: Qudrilterl KIT with digonl I, which is the perpendiculr isector of digonl TK nswers to ercises NSWRS TO XRISS 57

16 nswers to ercises Show: KIT is kite Tsk 4: To show tht qudrilterl is kite, you use the distnce formul to show tht only two pirs of djcent sides hve the sme length. Tsk 5: KI (0 ) ( 0) IT (0 )) ( ( 0) T (0 )) ( (c 0) c K ( ) 0 (0 c) c djcent sides KI nd IT hve the sme length nd djcent sides T nd K hve the sme length, nd ecuse c the pirs re not equl in length to ech other. Therefore, KIT is kite y definition. Therefore, if only one digonl of qudrilterl is the perpendiculr isector of the other digonl, then the qudrilterl is kite.. Tsk : Given: qudrilterl with midpoints connected to form second qudrilterl Show: The second qudrilterl is prllelogrm Tsk : y ( ) d, e G Q (0, 0) (d, e) (, 0 ) + d, c + e (, c) L ( ) U (, 0) +, c R ( ) Tsk : Given: Qudrilterl QU with midpoints, R, L, nd G Show: RLG is prllelogrm Tsk 4: To show tht qudrilterl is prllelogrm, we need to show tht opposite sides hve the sme slope. c c 0 c Tsk 5: slope of R e c e c e slope of RL d d d e c e c c slope of LG d d e e 0 e slope of G d d d Opposite sides R nd LG hve the sme slope, nd opposite sides RL nd G hve the sme slope. So they re prllel y the prllel slope property, nd RLG is prllelogrm y definition. Therefore the figure formed y connecting the midpoints of the sides of qudrilterl is prllelogrm.. Tsk : Given: n isosceles tringle with the midpoint of the se connected to the midpoint of ech leg, to form qudrilterl Show: The qudrilterl is rhomus Tsk : y Tsk : Given: Isosceles tringle with midpoint of se,, nd midpoints of legs, nd F, connected to form qudrilterl F. Show: F is rhomus Tsk 4: You need to show tht ll the sides of F hve the sme length. Tsk 5: y the distnce formul, h h h F h h h 0 h (, h ) h h h (, h) (0, 0) (, 0) F (, h ) (, 0) h F h 0 h h F F y the trnsitive property of equlity. Therefore, F is rhomus y the definition of rhomus. 58 NSWRS TO XRISS

17 HTR RVIW. Flse. The qudrilterl could e n isosceles trpezoid.. true. Flse. The figure could e n isosceles trpezoid or kite. 4. true 5. Flse. The ngles re supplementry ut not necessrily congruent. 6. Flse. See Lesson.5, mple. 7. true 8. perpendiculr 9. congruent 0. the center of the circle. four congruent tringles tht re similr to the originl tringle. n uiliry theorem proven specificlly to help prove other theorems. If segment joins the midpoints of the digonls of trpezoid, then it is prllel to the ses. 4. ngle isector 5. erpendiculr 6. ssume the opposite of wht you wnt to prove, then use vlid resoning to derive contrdiction. 7. Smoking is not glmorous. 7. If smoking were glmorous, then this smoker would look glmorous. This smoker does not look glmorous, therefore smoking is not glmorous. 8. Flse. The prllelogrm is rhomus. 9. Flse. ossile counteremple: 0. Flse; 60 so 90 only if (80 ) True (ecept in the specil cse of n isosceles right tringle, in which the segment is not defined ecuse the feet coincide). Given: Isosceles with ltitudes nd ; ; Show: y the Isosceles Tringle. y the definition of X ltitude nd y the Right ngles re ongruent. y the refleive property of congruence, so y S. y T. y the definition of congruence, nd, so y the sutrction property of equlity nd the Segment ddition. y the division property of equlity,,so divides the sides of proportionlly. Therefore y the onverse of the rllel/roportionlity.. true Given: Rhomus ROM with digonls RM nd O intersecting t R 4 O M Show: RM O ecuse digonls isect the ngles in rhomus, the digonls re ngle isectors. Sttement Reson.. Rhomus ngles. RO R. efinition of rhomus. RO R. efinition of congruence 4. R R 4. Refleive property of congruence 5. RO R 5. SS ongruence T 7. nd 4 re 7. efinition of liner liner pir pir 8. nd 4 re 8. Liner ir supplementry 9. nd 4 re right 9. ongruent nd ngles Supplementry 0. RM O 0. efinition of perpendiculr nswers to ercises NSWRS TO XRISS 59

18 nswers to ercises. true F y the Opposite ngles. m m m m y the definition of ngle isector. y the definition of prllelogrm. nd re supplementry y the Interior Supplements. nd re supplementry y the sustitution property. F y the onverse of the Interior Supplements. 4. Use the Inscried ngle, the ddition property, nd the distriutive property to get m m mn mt m mtn mt m m mn.ecuse there re 60 in circle, m m mn mt m T R ssume mh 45 nd mt 45. Use the Tringle Sum, the sustitution property, nd the sutrction property to get mh mt 90, which cretes contrdiction.therefore mh45 or mt Y T M Use the definition of midpoint, the Segment ddition, the sustitution property, nd the division property to get M Y TY nd YS YR. Then use the refleive property nd the SS Similrity to get MSY TRY. Therefore, MS TR y SST nd the multipliction property nd MS TR y ST nd the. 7. Y T S 4 Z O H R R Use the Line to etend ZO nd R. Then use the Line Intersection to lel s the intersection of ZO nd R. YR OR y the S ; thus Y O y T. Use the Tringle Midsegment nd the sustitution property to get TR (ZO Y). lso, use the Tringle Midsegment to get TR ZO. 8. The qudrilterl formed when the midpoints of the sides of rectngle re connected is rhomus. 8. G H Use the Right ngles re ongruent nd the SS ongruence to get H F GH GF. Then use T to prove tht FGH is rhomus. 9. The qudrilterl formed when the midpoints of the sides of rhomus re connected is rectngle. 9. I L O M X y the Tringle Midsegment oth M nd ON re prllel to LJ, nd oth O nd MN re prllel to IK. ecuse LJ nd IK re perpendiculr, we cn use corresponding ngles on prllel lines to prove tht the lines tht re djcent sides of the qudrilterl re lso perpendiculr. 0. The qudrilterl formed when the midpoints of the sides of kite re connected is rectngle. 0. y the Tringle Midsegment I M oth M nd ON re L prllel to LJ, nd oth O nd J MN re prllel to IK. ecuse LJ nd IK re perpendiculr, we cn use the to prove tht O N the lines tht re djcent sides of K the qudrilterl re lso perpendiculr.. Use the Line to construct chords nd. Then use the Inscried ngles Intercepting rcs nd the Similrity to get. Therefore, y SST nd the multipliction property,. K Y N J F O 60 NSWRS TO XRISS