Kha hc PenC N3 (Thy: L Anh Tun_Nguyn Thanh Tng) Hcmai.vn Hocmai.vn Ngi trng chung ca hc tr Vit 1 GI TNG CC BN 2 BI TON NHCNG 10 BI TON I KM Gii:- GiNl trung im caMA, khi :

23CK CGGKCN CM= = // MNhay GK // AB . -DoIl tm ng trn ngoi tip nnMI AB MI GK (1) -GiPl trung im caACv doABC Acn tiAnn:

/ / / / MP BC MK BCGI MKAG BC GI BC (2) -T (1) v (2) , suy raIl trc tm ca tam gicMGK(pcm). Phn tch hng gii: Vi kt qu Bi 9 ta c cIl trc tm ca tam gicMGK , t y ta d dng suy ra c ta im M ta imC(do3 MC MG = ) ta imA(doKl trng tmACM A ) ta imB (do M l trung im caAB ). Sau y l li gii chi tit cho bi ton: Gii:Gi ( )7 1;3 3( ; )3;KM x yM x yGM x y | |= | \ . = . Ta c : 1 1;3 31;03GIKI | |= | \ .| |=|\ . Theo kt qu Bi 9 ta cIl trc tm ca tam gicMGK( )1 7 1 10. 0 33 3 3 3(3;1)11 . 03 03x yGI KM xMyKI GMx | | | | + = || = = \ . \ . == = Bi 9. Cho tam gicABCcn tiAvMl trung im caAB . Gi, I Gln ltl tm ng trn ngoi tip, trng tm tam gicABC . Chng minh rngIl trc tm ca tam gicMGK. Bit rngKl trng tm ca tam gicACM . Bi 9.1. Trong mt phng ta Oxy , cho tam gicABCcn tiAvM l trung im caAB . Bit 8 1;3 3I | | |\ . l tm ng trn ngoi tip tam gicABCv(3;0) G , 7 1;3 3K | | |\ . ln lt l trng tm tam gic ABCvACM . Tm ta cc nh ca tam gicABC. Kha hc PenC N3 (Thy: L Anh Tun_Nguyn Thanh Tng) Hcmai.vn Hocmai.vn Ngi trng chung ca hc tr Vit 2 Mt khc G l trng tmABC Ann: ( )( )3 3 3 3 33 (3; 2)2 1 3 0 1C CC Cx xMC MG Cy y = = = = = DoKl trng tm tam gicACMnn: 3 ( ) 7 (3 3) 1(1;2)3 ( ) 1 ( 2 1) 2A K C MA K C Mx x x xAy y y y= + = + = = + = + = Ml trung im caAB , suy ra(5;0) B . Vy(1;2), (5;0), (3; 2) A B C . Phn tch hng gii: *) Do(2; 3) C v 7 1;3 3K | | |\ . l trng tm tam gicACMnn ta d dng suy ra c ta trung imNcaAM( vi 32CN CK = ). *) Vi kt qu ca Bi 9, ta c cCM KI ,t y ta s vit c phng trnhCM . *) imMc xut hin trong Bi ton 3, khiM CM evMK AI (viAtham s ha c theo im M doNl trung im caAM ) . Ngha l nh Bi ton 3, gip ta tm c ta imMvA . T y ta s suy ra ta imB( Ml trung im caAB ). Sau y l li gii chi tit cho bi ton. Gii:GiNl trung im caAM , khi : 3 42 212 3 3 11;12 2 3 23 322 3NNNNxxCN CK Nyy | | = = | \ . | |= |= \ . | | + = + |\ . Gi G l trng tm tam gicABC . Theo kt qu Bi 9 ta c: Il trc tm ca tam gicMGK KI MG hayKI CM .CMi qua(2; 3) C , nhn 1 1;0 (1;0)3 3KI| |= = |\ . lm vecto php tuyn nn CMc phng trnh:2 0 x =Bi 9.2. Trong mt phng ta Oxy , cho tam gicABCcn tiAv(2; 3) C . Bit 5 2;3 3I | | |\ . l tm ng trn ngoi tip tam gicABCv7 1;3 3K | | |\ . l trng tm tam gicACM , viM l trung im caAB . Tm ta cc nh cn li ca tam gicABC, bitAkhng trng vi gc ta . Kha hc PenC N3 (Thy: L Anh Tun_Nguyn Thanh Tng) Hcmai.vn Hocmai.vn Ngi trng chung ca hc tr Vit 3 Gi(2; ) (0;1 ) M m CM A m e (doNl trung im caAM ) Suy ra 5 5;3 32 2;3 3AI mMK m | |= | \ .| |= |\ . Do GI MK hay 2010 5 2. 0 0 01 9 3 3mAI MK AI MK m m m mm= | || | = + = = | | =\ .\ . Suy ra(0;1) Ahoc(0;0) A O (loi) VMl trung im caABnn(4; 1) B . Vy(0;1) Av(4; 1) B . Phn tch hng gii: *) Vi kt qu Bi 9 ta c cKI CM v suy ra c phng trnhCM . *) Lc ny ta thy xut hin Bi ton 3, khiM MC evMto vi, I Etam gic vung tiM . Ngha l nh Bi ton 3 s gip ta tm c ta imM .*) Khi c ta imMta svit c phng trnhAB(i qua, M E ) , ng thi tm c ta imP(32MP MK = ). Khi xut hin Bi ton 5, khi, A Cln lt thuc hai ng thng, AB CMv lin h vi imPqua h thc vecto (trong trng hp nyPl trung im caAC ). T y ta suy ra c ta im AvC . *) ViMl trung imAB , ta d dng suy ra c ta imB . Sau y l li gii chi tit cho bi ton: Gii:Gi G l trng tm tam gicABC .Theo kt qu Bi 9 ta cIl trc tm ca tam gicMGK

KI MG hayKI CM .CMi qua(2;0) Fv nhn 2 2;0 (1;0)3 3KI| |= = |\ . lm vecto php tuyn nn CMc phng trnh:2 0 x =Bi 9.3. Trong mt phng ta Oxy , cho tam gicABCcn tiAvMl trung im caAB . Bit 8 8;3 3I | | |\ . l tm ng trn ngoi tip tam gicABC , 10 8;3 3K | | |\ . l trng tm tam gicACM . Cc ng thng, AB CMln lt i qua cc im(0;3), (2;0) E F . Tm ta cc nh ca tam gicABCbitAc tung dng. Kha hc PenC N3 (Thy: L Anh Tun_Nguyn Thanh Tng) Hcmai.vn Hocmai.vn Ngi trng chung ca hc tr Vit 4 Gi ( )2 8;3 3(2; )2; 3IM mM m CMEM m | |= | \ . e = , khi : 2(2;4)44 8. 0 ( 3) 0 3 17 20 05 52; 3 33 3MmIM EM IM EM m m m mM m = | |

= + = + = | ||

=\ . |

\ . Vi(2;4) M , suy raABi qua(0;3), (2;4) E Mnn c phng trnh:2 6 0 x y + = . DoKl trng tm tam gicACMnn 3 102 24 2 3 3(4;2)2 2 3 84 42 3PPPPxxMP MK Pyy | | = | = \ .= =| | = |\ . Gi(2 6; ) A a a AB ev(2; ) C c CM e(vi0 a > ) Khi Pl trung im ca 2 2 6 2 8 6 (6;6)2 4 2 (2; 2)A C PA C Px x x a a AACy y y a c c C+ = + = = + = + = = VMl trung im ca( 2;2) AB B Vi 52;2M | | |\ ., suy raABi qua 5(0;3), 2;3E M | | |\ . nn c phng trnh:2 3 9 0 x y + = . DoKl trng tm tam gicACMnn 3 102 242 3 3 194;192 6 5 3 8 563 2 3 3PPPPxxMP MK Pyy | | = = | \ . | |= |= \ . | | = |\ . Gi(3 ;3 2 ) A a a AB ev(2; ) C c CM e(vi 32a ) Khi Pl trung im ca 3 2 8 2219 222 3 23 3A C PA C Pa ax x xACy y y a c c+ = = + = + = + = = (loi) Vy(6;6), ( 2;2), (2; 2) A B C . Bi 9.4. Trong mt phng ta Oxy , cho tam gicABCcn tiAvMl trung im caAB . ng thng CMc phng trnh3 0 y =v 2 7;3 3K | | |\ . l trng tm ca tam gicACM . ng thngABi qua im 1;42D| | |\ . . Tm ta cc nh ca tam gicABC , bit Mc tung dng vtm ng trn ngoi tip tam gicABCnm trn ng thng 2 4 0 x y + = . Kha hc PenC N3 (Thy: L Anh Tun_Nguyn Thanh Tng) Hcmai.vn Hocmai.vn Ngi trng chung ca hc tr Vit 5 Phn tch hng gii: *) Vi kt qu Bi 9 ta c cKI CM ( Il tm ng trn ngoi tipABC A ), cng viIthuc ng thng2 4 0 x y + =gip ta tm c ta imI . *) Lc ny ta thy xut hin Bi ton 3, khiM MC evMto vi, I D tam gic vung tiM . Ngha l nh Bi ton 3 s gip ta tm c ta imM .*) Khi c ta imMta svit c phng trnhAB(i qua, M D ) , ng thi tm c ta imP(32MP MK = ). Khi xut hin Bi ton 5, khi, A Cln lt thuc hai ng thng, AB CMv lin h vi imPqua h thc vecto (trong trng hp nyPl trung im caAC ). T y ta suy ra c ta im AvC . *) ViMl trung imAB , ta d dng suy ra c ta imB . Sau y l li gii chi tit cho bi ton: Gii:Gi G,Iln lt l trng tm, tm ng trn ngoi tip tam gicABC . Theo kt qu Bi 9 ta cIl trc tm ca tam gicMGK

KI MG hayKI CM . Khi KIi qua 2 7;3 3K | | |\ . vung gc vi: 3 CM y nn c phng trnh : 203x + = . Suy ra ta imIl nghim ca h : 220 2 83; 38 3 32 4 03xxIx y y = + = | | |\ . + = = Gi( ;3) M m CM e(vime),suy ra 2 1;3 3MI m| |= |\ . v 1;12MD m| |= |\ . 22 1 1 7. 0 0 03 2 3 6MI MD MI MD m m m m| || | = + + = + = | |\ .\ . 0 m =hoc 76m = (loi) Suy ra(0;3) M . Khi ABi qua 1;4 , (0;3)2D M| | |\ . nn c phng trnh:2 3 0 x y + = . DoKl trng tm tam gicACMnn 3 20 01 2 3 3( 1;2)2 2 3 73 32 3PPPPxxMP MK Pyy | | = | = \ .= =| | = |\ . Gi( ;3 2 ) A a a AB ev( ;3) C c CM e Khi Pl trung im ca 2 2 1 (1;1)2 3 2 3 4 3 ( 3;3)A C PA C Px x x a c a AACy y y a c C+ = + = = + = + = = VMl trung im ca( 1;5) AB B . Vy(1;1), ( 1;5), ( 3;3) A B C . Kha hc PenC N3 (Thy: L Anh Tun_Nguyn Thanh Tng) Hcmai.vn Hocmai.vn Ngi trng chung ca hc tr Vit 6 Phn tch hng gii: *) Vi kt qu Bi 9 ta c cKI CM ( Il tm ng trn ngoi tipABC A ), cng viIthuc ng thng2 4 7 0 x y + + =gip ta tm c ta imI . *) Lc ny ta nhn ra Bi ton 1, khiC CM ev 52CI R = = . Nn ta s d dng tm ra c ta imC .Ta tham s ha c imAthng qua imM CM e(doKl trng tmACM A ). Ngha l tip tc nh Bi ton 1 gip ta tm c ta imAvM . T y ta suy ra c ta imB . Sau y l li gii chi tit cho bi ton trn:Gii:Gi G,Iln lt l trng tm, tm ng trn ngoi tip tam gicABC . Theo kt qu Bi 9 ta cIl trc tm ca tam gicMGK

KI MG hayKI CM . Khi KIi qua 2 7;3 3K | | |\ .

v nhn(7;5)CMu = lm vecto php tuyn, nnKIc phng trnh:

2 77 5 0 7 5 7 03 3x y x y| | | |+ + = + = ||\ . \ .

Suy ra ta imIl nghim ca h 77 5 7 07 72;2 4 7 0 7 2 22xx yIx yy =+ = | | |+ + =\ .= Gi(4 7;5 ) C t t CM + e , khi : 25 252 2R IC IC = = =

2 221 7 257 5 74 42 0 02 2 2t t t t t| | | | + + + = + = = ||\ . \ .hoc 2137t = (loi). Suy ra(4;0) C . Bi 9.5. Trong mt phng ta Oxy , cho tam gicABCcn tiAvMl trung im caAB . ng thng CMc phng trnh 5 7 20 0 x y =v 11 7;6 6K | | |\ . l trng tm ca tam gicACM . ng trn ngoi tip tam gicABCc tm nm trn ng thng2 4 7 0 x y + + =v c bn knh bng 52. Tm ta cc nh ca tam gicABC , bitAv Cc ta nguyn . Kha hc PenC N3 (Thy: L Anh Tun_Nguyn Thanh Tng) Hcmai.vn Hocmai.vn Ngi trng chung ca hc tr Vit 7 Gi(4 7 ;5 ) M m m CM + e , khi Kl trng tm tam gicACMnn: 11 53 ( ) (4 7 4) 75 72 27 ; 57 7 2 23 ( ) (5 0) 52 2A K M cA K M cx x x x m mA m my y y y m m= + = + + = | | |\ .= + = + = ( ) ( )2 22 2 21(1; 1)2527 6 5 148 168 47 072 12; 47 237 3774AmIA R m m m mAm

=

= + + = + + = | |

| =

\ .

DoAc ta nguyn nn(1; 1) A 1 5; (0; 4)2 2M B| | |\ . (vMl trung im caAB ) Vy(1; 1), (0; 4), (4;0) A B C . -----*-----*-----*------*------*-------*-------*-------*--------*-------*--------*--------*-------*--------*-------*------- Gii:Cch 1 (dng hnh hc phng thun ty) Cch 1.1:

Gi{ } AC BD I = vEl trung im caDI .Khi MNEA l hnh bnh hnh(v, MA NEcng song song v bng 2CD) , suy raAE // MN (1) Mt khc :, NE DA DI AN nnEl trc tm tam gicDNA, suy raAE DN (2) T (1) v (2) suy raMN DN hay 090 MND =(*) Khi 0180 MND DAM + =nnMNDA ni tip ng trn, suy ra : 045 DMN DAN = = (cng chn DM ) (2*) T (*) v (2*) suy ra tam gicDMNvung cn tiN(pcm). Bi 10. Cho hnh vungABCD. GiMl trung im ca cnhABvNl im thuc onACsao cho3 AN NC = . Chng minh tam gicDMNvung cn. Kha hc PenC N3 (Thy: L Anh Tun_Nguyn Thanh Tng) Hcmai.vn Hocmai.vn Ngi trng chung ca hc tr Vit 8 Cch 1.2: Ging thng i quaNvung gc viAB , ct, AB CDln lt l, P Q.Khi MPN NQD A = A (c.g.c) 090 PNM QDN MNDDN MN= = = Suy ra tam gicDMNvung cn tiN(pcm). Cch 2 (dng h thc lng trong tam gic) t 2 3 2; ;2 4 4 4a AC a aAB a AM CN AN = = = = = . Khi : +) Xt tam gicADM ,ta c: 222 2 2 252 4a aDM AD AM a| |= + = + = |\ . (1) +) Xt tam gicAMN , ta c:

222 2 2 0 23 2 3 2 52 . .cos 2. . .cos452 4 2 4 8a a a aMN AM AN AM AN NAM a| || |= + = + = | | |\ .\ .(2) +) Xt tam gicDCN , ta c: 22 2 2 2 0 22 2 52 . .cos 2. . .cos454 4 8a aDN CD CN CDCN DCN a a a| |= + = + = | |\ . (3) T (1), (2), (3) ta c : 2 2 2MN DNMN DN DM= + =, suy ra tam gicDMNvung cn tiN(pcm). Cch 3 (dng phng php vecto)

tAB a = Ta c: ( )( )1 3 1 3 1 32 4 2 4 4 41 1 3 14 4 4 4MN MA AN DC AC DC AD DC DC DADN DC CN DC CA DC CD DA DC DA= + = + = + + = = + = + = + + = + Khi : 22 2 2 222 2 2 21 3 1 9 3 5.4 4 16 16 8 83 1 9 1 3 5.4 4 16 16 8 8MN DC DA DC DA DC DA aDN DC DA DC DA DC DA a| |= = + = | \ .| | = + = + + = |\ . MN DN = (1) Li c: ( )2 21 3 3 1 3 1. . 04 4 4 4 16 2MN DN DC DA DC DA DC DA DC DA| || |= + = = | |\ .\ . MN DN (1) T (1) v (2) suy ra tam gicDMNvung cn tiN(pcm). Kha hc PenC N3 (Thy: L Anh Tun_Nguyn Thanh Tng) Hcmai.vn Hocmai.vn Ngi trng chung ca hc tr Vit 9 Cch 4 (dng phng php ta )

Chn h trc ta Oxyvi(0;0) D O , (0; ), ( ;0) A a C aSuy ra;2aM a| | |\ . v 3;4 4a aN | | |\ . nn 3 3; , ;4 4 4 4a a a aMN DN| | | |= = ||\ . \ . Khi : 2 23 3. 016 16MN DN a a = = v 2 2 258MN DN a = =Suy ra tam gicDMNvung cn tiN(pcm). Gii:Theo kt quBi 10 , tam gicDMNvung cn tiNnn: 3 5 22. ( , ) 2. 12 2DN MN d N DM = = = =Suy ra, D Mthuc ng trn tm 5 2,2N| | | |\ . c phng trnh: 2 23 1 252 2 2x y| | | |+ + = ||\ . \ . Khi ta im, D Ml nghim ca h : 2 21 (1;3)3 1 253 (1; 2)2 2 21 (1; 2)1 02 (1;3)x Dy Mx yx Dxy M = | | | |= + + = || \ . \ . = = = . DoD c honh m nn ta c (1; 2)(1;3)DM GiIl giao im caACvBD ; G l giao im caACvDM Khi G l trng tm ca tam gicABD nn ta c: 11 1 3(1 )43 1;43 ( 2) 3(3 ) 33GGG GxxDM GM Gy y= = | |= | = = \ . Bi 10.1. Trong mt phng ta Oxy , cho hnh vungABCD. GiMl trung im ca cnhAB , 3 1;2 2N | | |\ . l im trn cnhACsao cho3 AN NC = . Xc nh ta cc nh ca hnh vungABCD, bit ng thngDMc phng trnh1 0 x =vD c tung m. Kha hc PenC N3 (Thy: L Anh Tun_Nguyn Thanh Tng) Hcmai.vn Hocmai.vn Ngi trng chung ca hc tr Vit 10 Mt khc: 2 2 2 4.3 3 3 9AG AI AN AN = = = Khi 4 313 9 2 4(3;2)2 9 4 4 13 9 2A AAAA Ax xxAG AN Ayy y | | = | = \ .= =| | = |\ . Suy ra( 1;4) B (do Ml trung im caAB ) (0;1) ( 3;0) I C ( doIl trung im caBD vAC ) Vy(3;2), ( 1;4), ( 3;0), (1; 2) A B C D . Ch :Ngoi cch trn ta c th tm im, D Mbng cch s dng Bi ton 6.2 nh sau: Gi(1; ) D d DM evi0 d < , suy ra 5 1;2 2ND d| |= |\ . . Ta c(0;1)DMu = Khi 2 21.2 2cos 22.5 12 2DMDMdNDuNDM dND ud= = = | | | |+ ||\ . \ . hoc3 d =(loi) Suy ra(1;3) D , suy ra phng trnh ng thng: 1 0 NM x y + + = Ta imMl nghim ca h 1 0 1(1; 2)1 0 2x xMx y y = = + + = =

Phn tch hng gii: *) Yu cu bi ton vit phng trnhCD, gip ta hng ti vic gn kt cc d kin tm cc yu t lin quan ti ng thngCD. Vic bi ton cho bit ta hai im(1;2) Mv(2; 1) N cng vi dkin 3 AN NC = , khin ta ngh ti vic tm ta imE( vi { } MN CD E = ) . iu ny hon ton c th lm c nh vo Bi ton 5.1 khi ta suy lun c3 MN NE = . *) Lc ny nu tm thm c mt im trnCD th coi nh bi ton gii quyt xong.NhBi ton 1ta s nghtivic tm imD. C thvi kt qu ca Bi 10 ta c c tam gicMND vung cn tiNnnD thuc ng thngND (vit c phng trnh) v cchNmt khong khng iMN ( DN MN = ). Nh vy bi ton chuyn v ng ni dung Bi ton 1 nn ta c li gii nh sau: Bi 10.2 ( Khi A, A1 2014).Trong mt phng vi h ta Oxy, cho hnh vungABCD c imMl trung im ca onABvNl im thuc onACsao cho3 AN NC = . Vit phng trnh ng thng CD, bit rng(1;2) Mv(2; 1) N . Kha hc PenC N3 (Thy: L Anh Tun_Nguyn Thanh Tng) Hcmai.vn Hocmai.vn Ngi trng chung ca hc tr Vit 11 Gii: Gi{ } MN CD E =vHl hnh chiu vung gc ca Mtrn CD. Khi theo Talet ta c3 3MN ANMN NENE NC= = = (*) +) Gi( ; ) E x ysuy ra( 2; 1) NE x y = +v vi(1; 3) MN =

Do 71 3( 2)(*)33 3( 1)2x xyy= = = += 7; 23E| | |\ .

+) Theo kt qu Bi 10 ta c tam gicMND vung cn tiN nn suy ra (1; 3)DNn MN = = . Khi phng trnhND:3 5 0 x y =+) DoD ND enn gi(3 5; ) D t t +. Khi (*)2 2 20 (5;0)(3 3) ( 1) 10 ( 1) 12 ( 1; 2)t Dt t tt D= + + + = + =

= ng thng CD i qua 7; 23E| | |\ . vD nnvi : +)(5;0) Dsuy ra CD c phng trnh : 3 4 15 0 x y = +)( 1; 2) D suy ra CD c phng trnh :2 y = hay2 0 y + =Vy phng trnh CD cn lp l 3 4 15 0 x y = hoc2 0 y + =Ch : Ngoi cch gii trn cc bn c th tham kho thm cch gii khc V d 2 trong Bi ton 6.1. Gii:Gi G l trng tm ca tam gicABD khi ta d dng ch rac : 58NGCG = . Suy ra 5 5 5( , ) . ( , ) .48 8 2d N DM d C DM = = = Gi( ; 3 4) N t t eAvi0 t < ,khi : 5 5 3( , ) 12 2 2d N DM t t = = = hoc 72t =(loi) Suy ra 3 1;2 2N | | |\ . Lc ny bi ton a v Bi 10.1 nn ta c kt qu cui cng:( 3;1) C hoc( 3;0) C . Bi 10.3. Trong mt phng ta Oxy , cho hnh vungABCD. GiMl trung im ca cnhAB ,Nthuc ng thng:3 4 0 x y A + + =vl im trn cnhACsao cho 14CN AC = . Bit phng trnh ng thng: 1 0 MD x = . Xc nh ta nh Cca hnh vungABCD, bit khong cch t Cn ng thng MDbng 4 vNc honh m. Kha hc PenC N3 (Thy: L Anh Tun_Nguyn Thanh Tng) Hcmai.vn Hocmai.vn Ngi trng chung ca hc tr Vit 12 Phn tch hng gii: *) Nhn thy cc d kin trong bi ton ch xoay quanh 3 im, , M N D. Do ta ngh ti vic lin kt cc im ny vi nhau bng cu hi 3 im ny liu c to nn tam gic c bit khng ?, v chng ta c c cu tr li nh vo Bi 10 khi tam gicMND vung cntiN . iu ny s gip ta vit c phng trnhDNv suy ra c ta imN . *) Vic tm imMcng kh d dng khi ta nhn thy,Mthuc Bi ton 1 ( M MN evMN DN = ) . Sau khi tm c imMta quay v ni dung ging nh Bi 10.1 . Do ta c li gii chi tit sau: Gii:Theo kt qu Bi 10 ta c tam gicMND vungtiN Do DNi qua(5;1) Dv nhn (1;3)MNu = lmvecto php tuyn. Khi phng trnhDN :3 8 0 x y + =Suy ra ta imNl nghim ca h : 3 4 0 2(2;2)3 8 0 2x y xNx y y = = + = = Gi( ;3 4) M t t MN evi 43t > , khi theo kt qu Bi 10 ta c tam gicMND cn tiN Do 2 2 2 2 2( 2) (3 6) 10 ( 2) 1 3 MN ND t t t t = + = = =hoc1 t =(loi), suy ra(3;5) M GiIl giao im caACvBD ; G l giao im caACvDM Khi G l trng tm ca tam gicABD nn ta c:

113 5 3(3 )11 1133 ;5 1 3(5 ) 11 3 33GGGGxxDM GM Gyy= = | |= | = \ .= Mt khc: 2 2 2 4.3 3 3 9AG AI AN AN = = = Khi ( )( )11 42543 9(5;5)11 4 5 923 9A AAAA Ax xxAG AN Ayy y = = = = = Suy ra(1;5) B(do Ml trung im caAB ) (3;3) (1;1) I C (doIl trung im caBD vAC ) Vy(5;5), (1;5), (1;1) A B C . Bi 10.4. Trong mt phng ta Oxy , cho hnh vungABCD c nh(5;1) D . GiMl trung im ca cnhAB ,Nl im trn cnhACsao cho 14CN AC = . Bit ng thng i qua hai imMvNc phng trnh 3 4 0 x y = . Xc nh ta cc nh cn li ca hnh vungABCD, bit imMc tung dng. Kha hc PenC N3 (Thy: L Anh Tun_Nguyn Thanh Tng) Hcmai.vn Hocmai.vn Ngi trng chung ca hc tr Vit 13 Gii:Theo kt qu ca Bi 10 ta c tam gicMND vung cn tiN. Suy raMD EF . Do MDc phng trnh:3 4 0 x y =Mt khc ta c: 21 3 1 03 (0;2) 321 3( 1 0)N NNNx xNE FE Nyy | | = = |= \ . = = Lc ny bi ton c ni dung ging Bi 10.1 do ta c kt qu :( 4;0) B .

CM N CC BN C TI LIU ! Nguyn Thanh Tng https://www.facebook.com/ThayTungToan Bi 10.5. Trong mt phng ta Oxy , cho hnh vungABCD . GiMl trung im ca cnhAB ,Nl im trn cnhACsao cho 14CN AC = . Bit(1; 1) E l trung im ca onDM . Tm ta nhB ,bit 2;03F | | |\ . l trng tm tam gicAMNv imMc honh m.