CHNGII: NGDNGCALNGGICTRONGHèNHHC Lnggiỏclmtcụngcmnhtrongtoỏnhc,núcngdngtronggiicỏcdngtoỏn khỏc,inhỡnhnhhỡnhhc,khosỏthms,chngminhbtngthc Cỏcbitpchng nychyunờuranhngvớdvsdngcụngclnggiỏcchngminhnhngbitpkhúv giithiucho cỏcbnmtsbitoỏncbit. Bi1:(nhlýStewart) Cho ABCDl1imtrờncnhBC.tAD=d,BD=m,DC=n.Khiútacúcụngthcsau: (gil h thcStewart): Gii: KngcaoAH xột2tam giỏcABD vACDv theonhlýhmscosin,ta cú: Nhõntngv(1) v(2)theothtvin vm ricngli,ta cú: Do nờnt(3)suyra: nh lýStewartchngminhxong. *M rng: 1.Stewart(1717ư1785)lnhtoỏn hcvthiờnvn hcngiScotland. 2.Nutrongh thcStewartxộtAD lngtrungtuynthỡ th thcStewartcú: (4)chớnhl hthc xỏcnhtrungtuynquenbittrongtam giỏc 3.NutronghthcStewartxộtADlphõngiỏc.Khiútheotớnhchtngphõngiỏctrongta cú: ThthcStewartcú: ư11 Chỳý rng: T(5)v(6)suyra: (7)chớnhl hthc xỏcnhngphõngiỏc Vy,hthcStewartltngquỏthúacahthcxỏcnhngtrungtuynvngphõngiỏc óquenbit. Bi2:Cho ABCgisDvEl2imtrờncnhBCsaocho ngtrũnnitip cỏc ABD v ACEtipxỳcvicnhBCtngng tiMv N.Chngminhrng: Gii: Tacú: Vyngthccnchngminhtngngvingthcsau: (*) t pdngnh lýhmssintrong cỏc ABDv ACE,ta cú: Trong ABEtheonh lýhms sin,ta cú: Tngt: ư12 Thay(3)vo(1) cú: Thay(4)vo(2) cú: Do nờnt(5)v (6)suyra: Trong ABDta cú: Tngt: Túsuyra: (8) T(1)v(2)suyra: pdngnh lýhmscosintrong cỏctamgicasABDv ACEtacú: Tacú: v theo(8)cú (11) Tngttacú: ư13 (12) Thay(11),(12)vo(1) cú: ( ) (13) T(9)v(13)cú (14) T(3)(4) v(14)suyra Haysaukhithay Tacú : (15) Thay(7)vo(15)cú: Hay (*) Vy(*)ỳng vliucnchngminh. Bi3:(nhlớhmscosthnhtvitgiỏc) ã=g AB=a,BC=b,CD ChotgiỏcliABCD,trongú AB = a, BC = b, CD = c, DA = d , ã ABC = b ,BCD =c,DA=d.Chngminhrng: d = a + b + c - ab cos b - 2bc cos g - ac cos( b +g ) Gii: GiK,LtngngltrungimACvBDvMltrungimBC(ChxộtkhiK L,tclkhi ABCDkhụngphilhỡnhbỡnhhnh,vỡnu ABCDlhỡnhbỡnh hnhthỡ b + g = 1800 a = c,b =d v ktluntrờnliuhinnhiờn) Cú2khnngxyra: 1)NuABkhụngsongsongviCD ã = ã Gis AB ầ CD = E => KML AED VitrnghpABctCDvphớatrờn,tacú: ã AED = 1800 - ộở(1800 - b ) + (1800 - g )ựỷ= b + g - 1800 KhiABctCBvphớadi,tacú: ã AED = 1800 - ( b +g ) Trongchaitrnghpucú: cos AED = - cos( b + g ) ư14 Trong D MKL,theonhlớhmssin,tacú: KL2 = MK + Ml - ML.MK cosKML a c a c KL = + +2 cos( b + g ) 4 2 a c ac => KL2 = + + cos ( b +g )(1) 4 2 TheocụngthcEulervitgiỏc,tacú: a + b + c + d - e - f ) ( 2) ( Vi e = AC , f =BD ,thay(2)vo(1): KL2 = a + b + c + d - e - f = a + b + ac cos ( b +g )(3) Liỏpdngnhlớhmscos,tacú: e = a + b - abcos b (4) f = c + b -2bc cos g (5) Thay(4)v(5)vo(3),tacú: d = e + f - b + 2accos( b + g ) = a + b + c - 2ab cos b - 2bc cos g + 2ac cos( b +g ) 2)NuAB//CD Khiú b + g =1800 Vyngthctngngvi: d = a + b + c - cos b ( ab - bc ) - 2ac d = a + b + c - 2b cos b (a - c ) -2ac ( 6) Thtvy,kAE//BC, theonhlớhmscostrong D AEDtacú: d = b + ( c - a ) - 2b ( c - a) cosg = b + c + a - 2b ( c - a )cos b -2ac Vy(6)ỳng.úchớnhlpcm. *Chỳý: 1.NhclicụngthcEulersauõy: ChotgiỏcliABCD,trongú AB = a, BC = b, CD = c, DA = d , AC = e,BD = f GiKvLl trungimACvBD.Khiútacú: KL2 = a + b + c + d - e - f ( ) ChngminhcụngthcEulernhsau: XộttamgiỏcALC,theotớnhchttrungtuyn: LC + 2LA2 - AC2 2 BC + 2CD - BD 2 AB + 2AD - BD2 + 2. - AC2 4 = = ( a + b + c + d - e - f ) KL2 = ư15 úlpcm. 2.Tacúcỏchgiikhỏcchobitoỏntrờnnhsau: Hinnhiờncú: uuur uuur uuur uuur AD = AB + BC + CD uuuruuur uuur uuur uuur uuur AD = AB + BC + CD + AB.BC + AB.CD + BC CD Theonhnghacatớchvụhngsuyra: d = a + b + c - ab cos b - 2bc cos g +2 ac cos ( AB,CD ) Do cos ( AB, CD ) = cos( b +g ) ( uuur uuur ) ( uuur uuur ) (Chỳýl cos AB, BC = cos (1800 - b ) = - cos b , cos BC , CD = cos (1800 - g )= - cos g =>pcm à, gi AH, AP, AM tng ng l ng cao, ng phõn giỏc >C Bi 4: Cho tam giỏc ABC cú B ã=a Chngminhrng: trongvngtrungtuynktA.t MAP tan A B-C =tan a cot 2 Gii: Cỏch1: MB = MC => S ABM = SACM 1 => c AM sin MAB = b AM sinMAC 2 ổA ổ A => c.sin ỗ + a ữ = b.sin ỗ - a ữ (1) ố2 ứ ố ứ Theonhlớhmssin,t(1)tacú: ổA ổ A sin C sin ỗ + a ữ = sin Bsinỗ - a ữ ố2 ứ ố ứ A A A A => sin C sin cos a + sin C cos sin a = sin B sin cos a - sin Bcos sina 2 2 A A => cos sin a ( sin B + sin c ) = sin cos a ( sin B - sinC) 2 A B+C B -C A B+C B - C => cos sin a sin cos = sin cos a cos sin 2 2 2 A B -C A B -C => sin a cos cos =cos a sin sin ( 2) 2 2 A B -C Chiac2vca(2)chocos cos a sin tacú: 2 A B-C tan =tan a cot 2 úlpcm. Cỏch2: ư16 ngphõngiỏctrongAPkộodictngtrũnngoitip D ABCtiI.KộodiOIctngtrũn tiJ. DdngthyrngPATMltgiỏcnitip. ã = PAM ã=a =>PJM ã= Mtkhỏc PIM B - C Túsuyra: tan a cot B - C PM MI MI IJ = = (1) JM PM MJ IJ Theo h thc lng tam giỏc vuụng, ta cú: ỡùMI IJ = IC2 ùợMJ IJ = IJ Vythayvo(1)tac: B - C ổ IC 2 tan a cot =ỗ ữ = tan IJC = tan a ố JC ứ úlpcm. Cỏch3: A B-C = tan a cot 2 A B-C B - C tan tan = tan a tan 2 B - C tan tana (1) = A B +C tan tan 2 ngthc tan Theonhlớhmstan,tacú B - C tan b - c = b +c tanB + C Vyt(1)suyra: B - C tana b - c = ( 2) A b +c tan tan 2a = tan a cot KộodiAbmtonBE=bưc.APkộodict ECtiK => AI ^ EC vIE=IC Tacú: IK tana IK = AI = ( 3) A EI EI tan AI DthyMI//BE(ngtrungbỡnhtrong D BEC) ư17