cac dang bt vecto lop 10
TRANSCRIPT
CHNG I : VECTO
CHNG I : VECTOA. Vecto cung phng, hai vecto bng nhau:Bai 1: Cho hnh bnh hanh ABCD co tam la O . Tm cac vect t 5 iem A, B, C , D , O
a) Bang vect ;
b) Co o dai bang ( (Bai 2 : Cho tam giac ABC. Ba im M,N va P ln lt la trung im AB, AC, BC. CMR:
; .Bai 3: Cho t giac ABCD, goi M, N, P, Q lan lt la trung iem AB, BC, CD, DA.
Chng minh : .
Bai 4: Cho tam giac ABC co trc tam H va O tam la ng tron ngoai tiep . Goi B la iem oi xng B qua O . Chng minh : .
Bai 5: Cho hnh bnh hanh ABCD . Dng . Chng minh
B. CHNG MINH NG THC VECTO:Bai 1: Cho 4 im bt ki M,N,P,Q . Chng minh cac ng thc sau:
a)
;b)
;
c)
;
Bai 2: Cho ngu giac ABCDE. Chng minh rng:
a)
;
b)
Bai 3: Cho 6 im M, N, P, Q, R, S. Chng minh:
a) .b).Bai 4: Cho 7 iem A ; B ; C ; D ; E ; F ; G . Chng minh rang :
a) + + = +
b) + + = + +
c) + + + = + +
d) - + - + - =
Bai 5: Cho hinh binh hanh ABCD, co tm O. CMR: .Bai 6: Cho ngu giac eu ABCDE tam O Chng minh :
Bai 7: Cho luc giac eu ABCDEF co tam la O . CMR :
a) +++++=
b) ++ =
c) ++ =
d) ++ = ++ ( M tuy y ).
Bai 8: Cho tam giac ABC ; ve ben ngoai cac hnh bnh hanh ABIF ; BCPQ ; CARS
Chng minh rang :
+ + =
Bai 9: cho t giac ABCD. Goi I, J ln lt la trung im AC va BD. Goi E la trung im I J . CMR: .Bai 10: Cho tam giac ABC vi M, N, P la trung im AB, BC, CA. CMR:
a);b);
c) .
Bai 11: Cho hinh thang ABCD ( ay ln DC, ay nho AB) goi E la trung im DB. CMR:
.
Bai 12: ( H thc trung im) Cho 2 im A va B.a) Cho M la trung im AB. CMR vi im I bt ki :
b) Vi N sao cho . CMR vi I bt ki :
c) Vi P sao cho . CMR vi I bt ki :
Bai 13: ( H thc trong tm) Cho tam giac ABC co trong tm G:
a) CMR: . Vi I bt ki : .
b) M thuc oan AG va MG = GA . CMR
c) Cho tam giac DEF co trong tm la G CMR:
+ .
+ Tim iu kin 2 tam giac co cung trong tm.
Bai 14: ( H thc hinh binh hanh) Cho hinh binh hanh ABCD tm O. CMR:
a) ;
b) vi I bt ki : .
C. MT S BAI TOAN LIN QUAN N DAI:Bai 1: Cho tam gic ABC l tam gic u cnh 2a. Tnh di cc vect
Bai 2: cho hinh thoi ABCD canh a. , goi O la giao im cua 2 ng cheo. Tinh:
|| ; ; .Bai 3: Cho hinh vung ABCD canh a. Tinh:
;
.
Bai 4: Cho t giac ABCD. Goi I, J la trung im cua AC va BD. Hay tinh :
.
D. Chng minh 3 im thng hang:Bai 1. Cho tam gic ABC v M, N ln lt l trung im AB, AC.
a) Gi P, Q l trung im MN v BC. CMR : A, P , Q thng hng.
b) Gi E, F tho mn : , . CMR : A, E, F thng hng.
Bai 2. Cho tam gic ABC, E l trung im AB v F thuc tho mn AF = 2FC.
a) Gi M l trung im BC v I l im tho mn 4EI = 3FI. CMR : A, M, I thng hng.
b) Ly N thuc BC sao cho BN = 2 NC v J thuc EF sao cho 2EJ = 3JF. CMR A, J, N thng hng.
c) Ly im K l trung im EF. Tm P thuc BC sao cho A, K, P thng hng.
Bai 3. Cho tam gic ABC v M, N, P l cc im tho mn : , , . CMR : M, N, P thng hng. ().
Bai 4. Cho tam gic ABC v L, M, N tho mn
EMBED Equation.DSMT4 , . CM : L, M, N thng hng.
Bai 5. Cho tam gic ABC vi G l trng tm. I, J tho mn : , .
a) CMR : M, N, J thng hng vi M, N l trung im AB v BC.
b) CMR J l trung im BI.
c) Gi E l im thuc AB v tho mn . Xc nh k C, E, J thng hng.
Bai 6. Cho tam gic ABC. I, J tho mn : . CMR : ng thng IJ i qua G.
Bai 7: Cho tam giac ABC co AM la trung tuyen. Goi I la trung iem AM va K la mot iem tren canh AC sao cho AK = AC. Chng minh ba iem B, I, K thang hangBai 8: Cho tam giac ABC. Hai iem M, N c xac nh bi cac he thc . Chng minh MN // AC.
E. Phn tich vecto theo cac vecto khac phng. Xc nh v tr mt im tho mn mt ng thc Vect:Bai 1: Cho 3 im A, B, C. Tm v tr im M sao cho :
a)
b)
c)
d)
e)
f)
Bai 2: Cho tam giac ABC co I, J , K lan lt la trung iem BC , CA , AB . G la trong tam tam giac ABC . D, E xac nh bi : = 2va =
. Tnh va theo va . Suy ra 3 iem D,G,E thang hang
F. Truc toa va h truc toa Bai 1 : Cho tam giac ABC . Cac iem M(1; 0) , N(2; 2) , p(-1;3) lan lt la trung iem cac canh BC, CA, AB. Tm toa o cac nh cua tam giac
Bai 2 : Cho A(1; 1); B(3; 2); C(m+4; 2m+1). Tm m e 3 iem A, B, C thang hang
Bai 3 : Cho tam giac eu ABC canh a . Chon he truc toa o (O; ; ), trong o O la trung
iem BC, cung hng vi , cung hng .
a) Tnh toa o cua cac nh cua tam giac ABC
b) Tm toa o trung iem E cua AC
c) Tm toa o tam ng tron ngoai tiep tam giac ABC
Bai 4 : Cho luc giac eu ABCDEF. Chon he truc toa o (O; ; ), trong o O la tam luc giac eu ,
cung hng vi , cung hng .
Tnh toa o cac nh luc giac eu , biet canh cua luc giac la 6 .
Bai 5:Cho A(-1; 2), B (3; -4), C(5; 0). Tm toa o iem D neu biet:
a) 2 + 3 =
b) 2 = 2 +
c) ABCD hnh bnh hanh
d) ABCD hnh thang co hai ay la BC, AD vi BC = 2AD
Bai 6 :Cho hai iem I(1; -3), J(-2; 4) chia oan AB thanh ba oan bang nhau AI = IJ = JB
a) Tm toa o cua A, B
b) Tm toa o cua iem I oi xng vi I qua B
c) Tm toa o cua C, D biet ABCD hnh bnh hanh tam K(5, -6)
Bai 7: Cho =(2; 1) ;=( 3 ; 4) va =(7; 2)
a) Tm toa o cua vect = 2 - 3 +
b) Tm toa o cua vect thoa + = -
Tm cac so m ; n thoa = m+ n
Bai 8 : Trong mt phng ta Oxy cho A(4 ; 0), B(8 ; 0), C(0 ; 4), D(0 ; 6), M(2 ; 3).
a/ Chng minh rng: B, C, M thng hng v A, D, M thng hng.
b/ Gi P, Q, R ln lt l trung im cc on thng OM, AC v BD. Chng minh rng: 3 im P, Q, R thng hng.
Bai 9. Trong mt phng ta Oxy cho hai im A(1 ; 3), B(-2 ; 2). ng thng i qua A, B ct Ox ti M v ct Oy ti N. Tnh din tch tam gic OMN.
Bai 10. Trong mt phng ta Oxy cho G(1 ; 2). Tm ta im A thuc Ox v B thuc Oy sao cho G l trng tm tam gic OAB.
Bai 11. Trong mt phng ta Oxy cho A(-4 ; 1), B(2 ; 4), C(2 ; -2).
a/ Chng minh A, B, C l 3 nh ca mt tam gic.
b/ Tnh chu vi ca tam gic ABC.
c/ Xc nh ta trng tm G v trc tm H.
Bai 12. Cho tam gic ABC vi A(1 ; 2), B(5 ; 2), C(1 ; -3).
a/ Xc nh ta im D sao cho ABCD l hnh bnh hnh.
b/ Xc nh ta im E i xng vi A qua B.
c/ Tm ta trng tm G ca tam gic ABC.
Bai 13. Cho A(1 ; 3), B(5 ; 1).
a/ Tm ta im I tha
b/ Tm trn trc honh im D sao cho gc ADB vung.
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