phần loại và phương pháp giải hình học 10

516.0076
PH121L
NGUYEN PHU KHANH
^I^^IKWli^^j
)AU THANH KY - NGUY§N MJNH NHIEN "
_JYEN ANH TRUING - NGUYEN TAN SIENG
D6 NGOG THUY
(Nhom giao vien chuyen toan trifdng THPT)
PHANLOAI
&
PHCONG PEIAP GlAl
mmm WHIM
(Tai Ban, SCfa ChOa Va Bo Sung)
• Danh cho hoc sinh Idp 10 on tap va nang cao kien thufc
• Bien soan theo noi dung sach giao khoa cua bp GD&DT
AT
BAN
DAI HOC QUOC GIA HA NOI
1
w
a\.
7«3UYElsrPFrtrKFwi\iFr
DAU
THANH KY
-
NGUYEN
MINH
NHIEN
NGUYEN
ANH TRUdNG
-

NGUYEN
TAN
SIENG
D6
NGOC
THUY
(Nhom
glao
vien
chuyen
toan trudng
THPT)
if
I!
ri;
:
PEIANLQAI
& FHlTdNG PHAP GIAI
(Tai Ban, Sufa ChOa
Va Bo
Sung)
Danh cho hoc sinh Idp 10 on tap va nang cao kien thiJc
Bien soan theo npi dung sach giao khoa cua bp
GD&DT
mm,
^ ^^ ^Allj
i/Sn riiA HA NOI
Cac em hoc sinh than men!.
"Phan
loai va

phuong
phap
giai
Hinh hoc 10" la mot trong nhi>ng cuon
thuoc
bo
sach
"Phan
loai
va
phuang
phap
giai
lop
10,11,12
", do nhom tac gia
chuyen toan THPT bien
soan.
Vai
each
vie't khoa hoc va sinh dpng, cuon
sach
se giup ban doc tiep can voi
mon
Toan
mot
each
tu nhien, khong ap luc, ban doc trd nen tu tin va nang
dong hon; hieu ro ban
chat,

biet
each
phan tich de tim ra trong tarn cua van de
va biet giai thich, lap luan cho tung bai toan. Su da dang cua he thong bai tap
va
tinh
huong giup ban doc iuon hung thu khi giai toan.
Tac gia chii trong bien
soan
nhung cau hoi ma, noi dung ca ban bam sat
sach
giao
khoa va cau true de thi Dai hoe, dong thai phan bai tap thanh cac
dang toan eo 161 giai chi tie't. Hien nay de thi Dai hoc khong kho, to hgp cua
nhieu van de don gian, nhung chiVa nhieu cau hoi ma neu khong nam
chac
ly
thuyet se lung tung trong
viec
tim loi giai bai toan. Vai mot bai toan, khong
nen
thoa
man
ngay
voi mot loi giai minh vua tim
dugc
ma phai
cogang
tim
nhieu

each
giai nhat cho bai toan dcS, moi mot
each
giai se c6 them phan kien
thue moi on tap.
Mon
Toan
la mot mon rat ua phong
each
tai tu, nhung phai la tai tu mot
each
sang
tao va thong minh. Khi giai mot bai toan, thay vi dCing thoi gian de
luc loi tri nha, thi ta can phai suy nghi phan tich de tim ra phuang phap giai
quyet bai toan do. Doi voi
Toan
hoc, khong c6 trang
sach
nao la thua. Tung
trang, tung dong deu phai hieu. Mon
Toan
doi hoi phai kien nhan va ben bi
ngay
tu nhung bai tap dan gian nhat, nhi>ng kien thue co ban nhat. Vi chinh
nhiing
kien thue co ban moi giup ban doe hieu
dugc
nhung kien thue nang cao
sau nay.
Ludwig

Van
Beethoven
tung noi: "Gigt
nuoe
co the lam mon tang da,
khong phai vi gigt
nuoe
co sue manh, ma do
nuae
chay
lien tuc
ngay
dem. Chi
CO
sir phaii da'u khong met moi moi dem lai tai nang. Do do ta co the
khSng
djnh,
khong nhich tung
buae
thi khong bao gio co the di xa ngan dam".
Mac dii tac gia da danh nhieu tam huyet cho cuon
sach,
song
su sai sot la
dieu kho tranh khoi. Chung toi rat mong nhan
dugc
su phan bien va gop y quy
bau ciia quy doe gia de nhung Ian tai ban sau cuon
sach
dugc

hoan
thi^n
han.
*' , r\ \ ^V'Q' ^^^y "^^^ nhom bien
soan
y\n Phu Khanh
Cty TNHH MTV
UVVHKmHg
VTgt
SAH4
Hinh
l.I
§1.
CACDINHNGHIA
A.
TOM TAT LY
THUYET
1.
Dinh
nghia
vecto:
Vccta
la doan thang co huong, nghia la
trong
hai diem mut cua doan thang da
chi ro diem nao la diem dau, diem nao la
diem cuoi.
Vecto
CO
diem dau la A, diem cuoi la B ta ki hieu: AB

Vecto
eon
dugc
ki hieu la: a, b, x, y,
Veeta
- khong la
vecto
co diem dau trung diem cuoi. Ki hieu la 0
2. Hai
vecto
cung
phuomg,
citng
huong.
- Duong thang di qua diem dau va diem cuoi ciia
vecto
ggi la gid cua
vecto
- Hai
vecto
co gia
song song
hoac
triing
nhau ggi la hai
vecto
ciing
phwang
- Hai
vecto

ciing phuang thi
hoac
eung huong
hoac
nguge
huong.
c
D
Hinh
1.2
H
Vi du: 6 hinh ve tren tren (hinh 2) thi hai
vecto
AB va CD ciing huong eon
EF va HG
nguge
huong.
Dac biet:
vecto
- khong cung huong voi mgi vec to.
3. Hai
vecto
bSng
nhau
A B
- Dg dai doan
thSng
AB ggi la do dai
vecto
AB,kihieu AB

Vgy AB =AB.
L
1
c
- Hai
vecta
hhng
nhau
neu chung ciing huong va eiing do dai.
Hinh
1.3
D
Vi du: (hinh 1.3) Cho hinh binh hanh ABCD khi do AB = CD
B. CAC
DANG
TOAN
VA
PHlTaNG
PHAP
GIAI.
DANG
TOAN
1: XAC DINH MOT VECTO; PHUONG, HUONG CUA
' VECTO; DQ DAI CUA VECTO
1. PHUONG PHAPGlAl.
• Xac djnh mot
vecto
va xac
dinh
su cung phuang, ciing huong ciia hai

vecto
theo
djnh nghla
• Dua vao cac
tinh
chat
hinh hoc cua cac hinh da cho biet de
tinh
dp dai cua
, mpt
vecto
2. CAC VI Dg. • I
Vi du 1: Cho tu
giac
ABCD. Co bao nhieu
vecto
khac
vecto-khong c6 diem
dau va diem cuoi la
dinh
cua tu
giac.
Lai
gidi
Hai
diem phan biet,
chang
han A, B ta xac djnh dupe hai
vecto
khac

vecto-
khong la AB, BA . Ma tu bon
dinh
A, B, C, D ciia tu
giac
ta c6 6 cap diem
phan biet do do c6 12
vecto
thoa man yeu cau bai toan.
Vi du 2: Chung minh rang ba diem A,B,C phan bi^t thang hang khi va chi
khi
AB, AC ciing phuong.
Lcri
giai
Neu A,B,C thang hang suy ra gia cua AB, AC deu la dudng thang di qua
ba diem A,B,C nen AB, AC cung phuong. i
Ngup-c lai neu AB, AC ciing phuong khi do duong
thSng
AB va AC
song
song
hoac
triing
nhau. Nhung hai duong thang nay ciing di qua diem A nen
hai duong thang AB va AC
triing
nhau hay ba diem A, B, C
thing
hang.
Vi du 3: Cho tam

giac
ABC. Goi M,N,P Ian
lupt
la trung diem cua
BC,CA,AB.
a) Xac djnh cac
vecto
khac
vecto
- khong ciing phuong v6i MN c6 diem dau
va diem cuoi lay trong diem da cho.
b) Xac
dinh
cac
vecto
khac
vecto
- khong ciing huong vdi AB c6 diem dau
va diem cuoi lay trong diem da cho.
c) Ve cac
vecto
bang
vecto
NP ma c6 diem dau A, B .
Lcri
gidi
(Hinh
1.4)
a) Cac
vecto

khac
vecto
khong ciing phuong voi MN la
NM,
AB, BA, AP, PA, BP, PB.
b) Cac
vecto
khac
vecto
- khong cung huong voi AB la AP, PB, NM .
c) Tren tia CB lay diem B' sao cho BB' = NP ^ ' ^
Khi
do ta
CO
BB' la
vecto
c6 diem
dau la B va bang
vecto
NP. '
Qua A dung duong thang
song
song
voi duong thang NP.
Tren duomg thang do lay diem
A'
sao cho AA" ciing huong
vo-iNP
va
AA"

= NP.
Hinh
1.1"
Khi
do ta
CO
AA'
la
vecto
c6 diem dau la A va bang
vecto
NP.
Vi du 4: Cho hinh vuong ABCD tam O
canh
a . Goi M la trung diem cua
AB,
N la diem doi xung voi C qua D. Hay
tinh
do dai ciia
vecto
sau:
MD,
MN . ^
Lai
gidi
(hinh 1.5)
Ap
dung djnh ly
Pitago
trong tam

giac
vuong MAD ta c6
DM^
=
AM^
+
AD^
=
Suy ra
MD
= MD-
a'
2 5a'
+ a = —
Qua N ke duong thing
song
song
vol AD cat AB tai P. |
Khi
do tu
giac
ADNP la hinh vuong va|
a 3a
PM
= PA + AM - a + - = —.
-4 M
Ap
dung djnh ly
Pitago
trong tam

giac
vuong NPM ta c6
MN^
=NP2+PM^ =a^ +
ax/TS
3a^
2 .
13a^
.MN
=
iVl3
Suy ra
MN
= MN =
•
3. BAI TAP
LUY£N
TAP. *^ ""^
Bai 1.1: Cho
ngii
giac
ABCDE . Co bao nhieu
vecto
khac
vecto-khong c6 diem
dau va diem cuoi la
dinh
ciia
ngii
giac.

^
Huang
dan
gidi
I Hai diem phan biet,
chang
han A, B ta xac
dinh
dupe hai
vecto
khac
vecto-
khong la AB, BA. Ma tir nam
dinh
A, B, C, D, E ciia ngu
giac
ta c6 10 ca
I
diem phan biet do do c6 20
vecto
thoa man yeu cau bai toan.
P
Bai
1.2:
Cho
ba
diem A,
B, C
phan bi^t thang hang.
a) Khi nao thl hai vecto

AB va AC
cung huong
? ^.
b) Khi nao thi hai vecto
AB va AC
ngiroc huong
?
Huang
dan
gidi
^
a)
A
nam ngoai doan
BC
b)
A
nam trong doan
BC
Bai
1.3:
Cho bon diem A,
B, C, D
phan biet.
a) Neu AB
= BC
thi
c6
nhan xet gi
ve ba

diem A,
B, C
f>r{>
b) Neu AB
=
DC thi
co
nhan xet gi
ve
bon diem A,
B, C, D
Huang
dan
gidi
,
(^fy j
a)
B la
trung
diem
cua
AC
b) A,
B, C, D
thang hang
hoac
ABCD
la
hinh
binh hanh,

hinh
thoi,
hinh
vuong,
hinh
chij' nhat
Bai
1.4:
Cho
hinh
thoi
ABCD
c6
tam
O.
Hay
cho
biet khang djnh nao sau
day
dung
?
^
b)
AB
=
DC
c)
OA
=
-OC

a) AB
=
BC
d)
OB
=
OA
e)
AB BC
BD
f)
2
OA
Huang
dan
gidi
a)
Sai b)
Dung
c)
Dung
d)
Sai e)
Dung f)
Sai
Bai
1.5:
Cho luc
giac
deu ABCDEF tam

O.
Hay tim
cac
vecto
khac
vecto-khong
CO
diem dau, diem cuoi
la
dinh
ciia luc
giac
va tam
O sao
cho
a)
Bang
voi AB
b)
Ngugc huong voi
OC
Huang
dan
gidi
a) Fd,OC,ED
b) c6,OF,BA,DE
Bai
1.6:
Cho
hinh

vuong ABCD canh a
,
tam
O va M la
trung
diem AB.
Tinh
dp dai cua
cac
vecto AB,
AC,
OA,
OM,
OA
+
OB.
Huang
dan
gidi
Ta CO
AB = AB
= a;
OA
=OA=-AC=
2
AC
= AC
=
\/ABVBC^
=

aV2
OM
= OM
= -
Ggi
E la
diem
sao
cho tu
giac
OBEA
la
hinh
binh
hanh khi do no cung
la
hinh
vuong
Ta CO OA
+
OB
=
OE
=>
OA
+
OB
=
OE
=

AB
= a
Hinh
1.40
Bai
1.7: Cho tam
giac
ABC deu
canh
a va G la
trong tam.
Goi I la
trung
diem
cua AG.
Tinh
do dai
cua cac
vecto AB, AG,
BI.
1
Huang
dan
gidi
S.
Ta CO
AB
= AB
= a
Goi

M la
trung
diem
cua BC
Ta CO
AG
-AM
=
-\//
3
3
= AG
=
-AM
=
-7AB2-BM2
=^
BI
—j_(aW^aV2T
= BI
=
N/BM'+MI2
=^^ +
^
=
—
Bai
1.8:
Cho truoc hai diem
A,B

phan biet. Tim tap hop
cac
diem
M
thoa man
Huang
dan
gidi
o MA
=
MB
=>
Tap hop diem
M la
duong
trung
true
cua
doan
MA
MB
MA
MB
thang
AB
DANG
TOAN
2:
CHUNG
MINH

HAI
VECTO
BANG
NHAU.
:
1.
PHLTONG
PHAPGlAl.
'
•
De
chiing
minh
hai
vecto bang nhau
ta
chung
minh
chung
c6
ciing
do
dai
va cung huong
hoac
dua vao nhan xet neu tii
giac
ABCD
la
hinh

binh hanh
thi
AB
=
DC
va AD = BC
2.
CAC
VI DU. .y
;,rt
7. ,
Vid«l:Ch^
giac
ABCD. Goi M, N,
P, Q
Ian
lugt
la
trung
diem AB,
BC,
CD,
DA. Chung
minh
rSng MN=QP
•
'
•
Lai
gidi

(hinh
1.6)
Do M,
N Ian
luat
la
trung
diem
cua AB va BC
nen MN
la
duong
trung
binh cua tam
giac
ABC
Suy
ra
MN//AC
va
MN
=
iAC
(1).
Tuong tu QP
la
duong
trung
binh
cua

tam
giac
ADC
suy
ra
QP//AC
va
QP
=
^AC
(2).
Tir (1) va (2) suy ra MN//QP va MN = QP
do do tur
giac
MNPQ la hinh binh
hanh
Vaytac6MN=QP
Vi du 2: Cho tarn
giac
ABC c6 trong tarn G . Goi I la trung diem
ciia
BC .
Dung diem B'
saocho
B'B = AG. ^
,>^,.,i
i,,
a) Chung minh rang BI = IC % -; • '''-^
b) Goi J la trung diem
ciia

BB'. Chung minh rang BJ = IG.
^
Loi^ax
(hinh 1.7)
a) Vi I la trung diem cua BC nen BI = CI va BI ciing Huang voi IC do do
hai
vecto
BI, IC
bang
nhau
hay BI = IC .
b)
Taco
B'B = AG suy ra B"B = AG va
BB7/AG.
Do do BJ, IG
cung
huong (1).
Vi
G la trong tam tam
giac
ABC nen
* IG = ^AG, J la trung diem BB' suy ra
BJ=1BB-
^ Vivay BJ = IG (2)
I
Tu (1) va (2) ta CO BJ = IG.
Hinh
1.7
Vi du 3: Cho hinh binh

hanh
ABCD.
Tren
cac
doan
thSng
DC, AB
theo
thu
hr lay cac diem M, N sao cho DM = BN. Goi P la
giao
diem
ciia
AM, DB
va Q la
giao
diem cua CN, DB. Chung minh rang AM = NC va
DP = QB.
LOT
^«» (hinh 1.8)
Ta CO DM = BN => AN = MC, mat
khac
AN
song song
voi MC
do do tu
giac
ANCM la hinh binh
hanh
v

SuyraAM = NC.
Xet tam
giac
ADMP va ABNQ ta c6
DM
= NB (gia thiet), _
PDM = QBN (so le
trong)
m
Mat
khac
DPM = APB (do! dinh)
va APQ = NQB (hai goc dong vj) suy ra DMP = BNQ.
Do do ADMP = ABNQ
(c.g.c)
suy ra DP = QB .
De
thay
DP, QB ciing huong vi vay DP = QB.
8
3. BAI TAP
LUYCN
TAP.
Bai 1.9: Cho tu
giac
ABCD
. Goi M, N, P, Q Ian lugt la trung diem AB, BC, CD,
DA. Chung minh rang MQ = NP .
m-ofi
Huong

dan
gidi
•
V^**^
Do M, Q Ian lugft la trung diem
ciia
AB va AD nen A
MQ la duong trung binh
ciia
tam
giac
ABD suy
ra
MQ//BD
va MQ = ^BD (1).
Tuong
tu NP la duang trung binh
ciia
tam
giac
CBD
suyra
NP//BD
va NP = iBD (2).
Tir (1) va (2) suy ra MQ//NP va NP = MQ do do tir
giac
MNPQ la hinh
binh
hanh
Vay

taco
MQ=NP .
Bai 1.10: Cho hinh binh
hanh
ABCD. Goi M, N Ian lugt la trung diem cua DC,
AB; P la
giao
diem
ciia
AM, DB va Q la
giao
diem
ciia
CN, DB . Chung
minh rang DM = NB va DP PQ = QB .
Huang
dan
gidi
'
Ta
CO
tu
giac
DMBN la hinh binh
hanh
vi DM = NB = | AB, DM
/
/NB.
Suy ra DM = NB.
Xet tam

giac
CDQ c6 M la trung diem
ciia
DC va
MP//QC
do do P la trung diem
ciia
DQ.
Tuong
tu xet tam
giac
ABP suy ra
dugc
Q la trung diem
ciia
PB
Vi
vay DP = PQ = QB tu do suy ra DP = PQ = QB
Bai 1.11: Cho hinh
thang
ABCD
c6 hai day la AB va CD voi AB - 2CD. Tu C
ve Ci = DA. Chung minh rang
^ b) Ai = iB = DC V
Huang
dan
gidi
a) AD = IC va DI = CB
a) Ta
CO

CI = DA suy ra AICD la hinh binh
hanh
=>AD = IC
Ta CO DC = AI ma AB - 2CD do do AI =
i
AB
=> I la trung diem AB
Ta
CO
DC = IB va DC / /IB => tir
giac
BCD! la hinh binh hanh
Suy ra DI = CB
b) I la trung diem cua AB=>AI = IB va tu
giac
BCDI la hinh binh hanh
=> IB = DC suy ra AI = IB = DC
Bai 1.12: Cho tam
giac
ABC c6 true tarn H va O tarn la duong tron ngoai tiep .
Goi B' la diem doi xung B qua O. Chung minh: AH =B'C .
Huang
dan
gidi
Taco
B'CIBC,
AH 1 BC ^ B'C
//AH
, B'AIBA, CH 1 AB => B'A //CH
Suy ra AHCB' la hinh binh hanh do do AH =B'C .

§2 TONG VA HIEUHAIVECTO
A. TOM TAT LY
THUYET
1.
Tonghaivecta
a) Dinh nghla: Cho hai
vecto
a; b. Tir diem A tuy y ve AB = a roi tu B ve
BC = b khi do
vecto
AC duoc goi la tong cua hai
vecto
a; b.
Ki
hieu AC = a + b (Hinh 1.9)
b) Tinh
chat:
+
Giac
hoan : a + b = b + a
+ Ket hop : (a + b) + c = a + (b + c)
+ Tinh
chat
vecto
- khong: a + 0 = a, Va
2.
Hieu
hai
vecta
Hinh

1.9 ?
a)
Vecto
doi cua mot
vecto.
Vecta
doi
ciia
vecto
a la
vecto
ngugc
huang va cung do dai vol
vecto
a
Ki
hieu -a
Nhu
vay a + (-a) = 0, Va va AB = -BA
b) Djnh nghta hieu hai
vecto:
Hieu
ciia hai
vecta
a va b la tong ciia
vecto
a va
vecto
doi ciia
vecto

b. Ki
hi^u
la a-b = a + (-b)
3. Cac quy tic:
Quy tac ba diem: Cho A, B ,C tuy y, ta c6 : AB + BC = AC
Quy tic hinh binh hanh: Neu ABCD la hinh binh hanh thi AB + AD = AC
Quy tac ve hieu
vecto:
Cho O, A, B
tiiy
y ta c6: OB -OA = AB
10
Chu ij: Ta c6 the mo rpng quy tic ba diem cho n diem A,,A2, ,An thi
A1A2 +
A2A3"
+ +
A„_,An
=
A,
A„
B. CAC
DANG
TOAN
VA
PHl/QNG
PHAP
GIAI.
DANG
TOAN
1: XAC DINH DO DAI TONG, HIEU CUA CAC VECTO.

1.
PHLTONG
PHAP
GIAI.
De xac
dinh
do dai tong hieu ciia cac
vecto
• Truoc tien sir dung djnh nghla ve tong, hieu hai
vecto
va cac
tinh
chat,
quy
tac de xac djnh phep toan
vecto
do.
• Dua vao h'nh
chat
cua hinh, sir dung djnh li Pitago, he thuc lugng trong tam
giac
vuong de xac djnh do dai
vecto
do.
2. CAC
ViDU.
Vidu
1: Cho tam
giac
ABC vuong tai A c6 ABC = 30" va BC =

aVS
.
Tinh
do dai ciia cac
vecto
AB + BC, AC - BC va AB + AC .
Lai
gidi
(hinh 1.10)
Theo
quy tac ba diem ta c6
. AB + BC = AC
AC
Ma sin ABC
^
BC
AC
=
BC.sin
ABC =
ax/5.sin30°
aVS
Do do
AB + BC = AC
= AC
•
AC-BC=AC+CB=AB
Ta CO AC^ + AB^ = BC^ ^ AB = >/BC^ - AC^ =
a>/T5
Vi

vay
AC-BC
AB =AB =
Goi D la diem sao cho tir
giac
ABDC la hinh binh hanh.
Khi
do
theo
quy tic hinh binh hanh ta c6 AB + AC = AD
Vi
tam
giac
ABC vuong 6 A nen tir
giac
ABDC la hinh chu nhat suy ra
AD
= BC = a>/5
Vay
AB + AC
AD
= AD = a>/5
11
Vtdu
2: Cho
hinh
vuong ABCD c6 tarn la O va canh a. M la mpt diem bat ky.
a) Tinh
AB +AD
OA

- BO
CD - DA
b) Chung
minh
rang u = MA + MB - MC - MD khong phu thuoc vi tri diem
M
. Tinh do dai vecto u
Lai
gidi
{hinh
1.11)
a) +
Theo
quy tac
hinh
binh hanh ta c6 AB + AD = AC
Suy ra
AB
+ AD
AC
= AC.
Ap
dung
dinh
li Pitago ta c6
AC^
= AB^ + BC^ = 2a^ ^ AC = >/2a
Vay
AB
+

AD
=
ayfl
+ Vi O la tam ciia
hinh
vuong nen OA = CO
suy ra OA - CB - CO - BO - CB
Vay
OA
- BO CB
= a
Hinh 1.11
+ Do ABCD la
hinh
vuong nen CD =
BA
suy ra CD - DA = BA + AD = BE) •
h Ma
|BD|
= BD = 7AB^ +
AD^
= suy ra
|CD
-
DA|
= a^f2
b)
Theo
quy tSc phep tru ta c6: u =
(MA

- MC) + (MB -
MD)
=
CA
+ DB
Suy ra u khong phu thuoc vj tri diem M .
Qua A ke duong thang song song v6i DB cat BC tai C .
Khi
do tu
giac
ADBC la
hinh
binh hanh (vi co cap canh doi song
song)
suy
ra DB = AC'
Do do u -
CA
+ AC' = CC'
Vi
vay
Icci = BC + BC' = a + a-2a
3. BAI TAP
LUY^N
TAP. ^
Bai 1.13: Cho tam
giac
ABC deu canh a. Tinh do dai cua cac
vecta
sau

AB-AC, AB + AC. ^
^
Huong
dan
gidi
'^A'
Theo
quy tac phep tru ta c6
AB-AC
= CB:
AB-AC
=BC = a
Gpi
A' la
dinh
ciia
hinh
binh hanh ABAC va
O
la tam
hinh
binh hanh do.
Hinh 1.45
Khi
do ta CO AB + AC = AA'.
Ta CO AO = yf^.
AB'-OB'=Ja'-^
Suy ra
• •••• ' /
AB

+ AC = AA' = 2AO = aS
Bai 1.14: Cho
hinh
vuong
ABCD
c6 tam la O va canh a. M la mot diem bat ky.
a) Tinh AB + 0D|,
AB-OC
+ OD
b) Tinh do dai vecto
M
A - MB - MC + MD
Huang
dan
gidi
a) Ta CO OD = BO => AB + OD = AB + BO = AO
AB + OD
= AO =
AC
ayfl
B'
Ta
CO OC = AO suy ra
1 1
V
=
6
B
= 0
Hinh 1.46

=>
AB-OC
+ OD
b) Ap dung quy tac phep tru ta c6
MA
-
MB
- MC + MD =
(MA
-
MB)
^
(MC
-
MD)
= BA - DC
LayB' la diem dol xung cua B qua A
do
-DC
= AB' BA -
DC
=
BA
+ Ais^BB'
Suy ra
MA-MB-MC
+ MD
BB'
= BB' = 2a
Bai 1.15: Cho

hinh
thoi
ABCD canh a va BCD = 60". Goi O la tam
hinh
thoi.
Tinh
AB + AD
OB-DC
Ta CO
AB
+ AD
OB-DC
AC
CO
Huang
dan
gidi
=
2acos30"
=aS,
0 a>/3
=
a
cos 30 =
OB+AC-OA
Bai 1.16: Cho bon diem A, B, C, O phan biet c6 do dai ba vecto OA, OB, OC
cung bang a va OA + OB + OC = 0
a) Tinh cac goc AOB, BOC, COA b) Tinh
Huang
dan

gidi
a) Tu gia thie't suy ra ba diem A, B, C tao thanh tam
giac
deu nhan O lam
trongtamdodo AOB = BOC=COA = 120" ,
b)
Gpi I la trung diem
BC.
Theo
cau a)
AABC
deu nen AI = — a
2
OB
+
AC-OA"
=a>/3
Bai 1.17: Cho goc Oxy. Tren Ox, Oy lay hai diem A, B . Tim dieu ki?n cua A,B
sao cho OA + OB nam tren phan
giac
ciia goc Oxy. !
Huang
dan
gidi
Dung
hinh binh hanh OACB. Khi do: OA+OB = OC
Vay OC nam tren phan
giac
goc xOy <=> OACB la hinh thoi <=> OA = OB.
DANG

TOAN
2: CHUNG MINHDANG THUC VECTO.
1. PHLTONG PHAPGIAI.
• De chung minh dang thuc
vecta
ta c6 cac
each
bien doi: venay thanh vekia,
bien doi tuong duong, bien doi hai veciing bang mot dai lirong trung gian.
Trong qua
trinh
bien doi ta can su dung
linh
boat
ba quy t3c
ti'nh
vecto.
Lmi
I/:
Khi bien doi can phai
liitmi<^
dicli,
chang han bien doi vc'phai, ta can
xem vetrai c6 dai luong nao de tu do lien tuang den kie'n thuc da c6 de lam
sao xuat hien cac dai lugng 6 vetrai. Va ta thuang bien doi ve'phuc tap ve
vedan gian hon.
2. CAC Vi DU.
Vtdu
1: Cho nam diem A,B,C,D,E . Chung minh r5ng
.aMB +CD + EA = CB + ED b) AC +

CD-EC
= AE-DB + CB
Lai
gidi
a) Bien doi vetrai ta c6 ^
VT = (AC + CB) + Cb + (Eb + DA)
= (CB + Eb) + (AC + Cb) + DA =(CB + Eb) + AD + DA
= CB + ED = VP (DPCM) : .
b) Dang thuc tuong duong voi
(AC-AE) +
(CD-CB)-EC
+ DB =
d<:>EC
+
BD-EC
+ DB = 0
BP + DB = 6 (dung) (DPCM).
Vidu
2: Cho hinh binh hanh ABCD tam O. M la mot diem bat ki trong m|it
phang. Chung minh rang .
a) BA + DA + AC = d
c) MA + MC = MB + MD .
b) OA + OB + OC + OD^d
LaigidHHinh
1.12)
a) Ta CO BA + DA + AC = -AB - AD + AC
= -(AB + AD) + AC ^
Theo
quy tac hinh binh hanh ta c6
AB + AD = AC suy ra

. ^ _. v^,? n
BA + DA + AC = -AC + AC = 0
b) Vi ABCD la hinh binh hanh nen ta c6:
OA
=
Cd=>OA
+ OC = OA +Ad = d
M.n,,,,,^
,
Tuong tu: OB + OD = 0 => OA + OB + OC + OD = 0 .
c)
Cach
1: Vi ABCD la hinh binh hanh nen AB = DC => BA + DC = BA + AB = 0
MA
+ MC = MB + BA + MD + DC
= MB + Mb + BA + DC = MB + Mb
Cach
2: Dang thuc tuong duang voi
Ac;
MA
- MB = MD - MC » BA - CD (dung do
ABCD
la hinh binh hanh)
Vi du 3: Cho tam
giac
ABC. Goi M, N, P Ian lugt la trung diem cua
BC, CA, AB . Chung minh rang . .
a) BM + CN +AP = d
b) AP +
AN-AC

+ BM = d
c) OA + OB + OC = OM + OIV + OP voi O la diem bat
ki.
'
Lai
gidi
{H\nh 1.13)
a) Vi
PN,MN
la duang trung binh cua tam
giac
ABC nen
PN//BM,
MN//BP
suy ra tu
giac
BMNP la hinh binh hanh
BM
= PN
N
la trung diem cua
AC
=>
CN
= NA
Do do
theo
quy tac ba diem ta co
BM
+ CN + AP = (PN + NA) + AP

= PA + AP = d
b) Vi tu
giac
APMN la hinh binh hanh
nen
theo
quy tMc hinh binh hanh ta c6
AP
+ AN = AM , ket hop voi quy t^c phep tru
=> AP + AN - AC + BM = AM - AC + BM = CM + BM
Ma
CM
+ BM = d do M la trung diem ciia BC.
Vay AP +
AN-AC
+
BM-d.
c) Theo quy tac ba
diem
ta c6
OA
+
OB
+
OC
=
(OP + PA) +
(OM
+
MB)

+
(ON +
NC)
= (OM
+
ON
+
OP)
+
PA
+
MB
+
NC
,
=
(OM+ON+OP)
-
(BM
+
CN
+
AP)
•
Theocau
a) ta c6 BM +
CN
+ AP =
d
suy ra OA

+
OB + OC =
OM +
^+^.
3. BAI TAP
LUY^N
TAP.
Bai
1.18: Cho bon
diem A,
B,C,
D. Chung minh
ring
a)
DA-CA
=
pB-CB
b)
AC
+
DA +
BD
= AD-CD+BA
„
' • Huang dan gidi
a) Ap
dung
quy tac
phep tru
ta c6 _

DA-CA
=
DB-CB<»DA-DB
=
CA-dB
c>BA = BA
(dung)
V
b)
Ap
dung
quy tac ba
diem
ta c6
AC
+
DA +
BD
= AD
-
CD
+
BA »
(DA
+ AC)
+
BD
= (BA
+
AD)

-
CD
o
DC + BD = 150 -
CD (diing)
Bai
1.19: Cho cac
diem
A, B,
C, D,
E, F .
Chung minh rang AD +
BE
+
CF = AE
+
BF
+
CD
Hu&ng
dan gidi
Cdch
1:
Dang thuc
can
chung minh tuang duong voi
( AE)
-
AE)
+ (BE -

BF)
+
(CF -
CD)
= d
o
ED + FE + DF = 6 o
EF
+
FF =
d (dung)
Cdch
2: VT =
AD
+ BE
+
CF = (AE +
ED)
+ (BF
+
FE)
+
(CD +
DF)
= AE +
BF
+
CC>
+
EC>

+
FE
+
DF
•
=
AE
+
BF
+
CC) =
VP
Bai
1.20: Cho
hinh binh hanh ABCD
tam O. M la mot
diem
bat ki
trong
mat
phang. Chung minh r^ng
a) AB +
OD
+
OC
= AC b) BA +
BC
+ 0B = 0D
c)
BA

+ BC
+
OB
=
MO-MB
Huang dan gidi
a) Ta
CO
OD = BO do do
AB
+
OE)
+
OC =
AB +
B6
+
OC
= Ad
+
OC
=
AC
b)
Theo quy tac
hinh binh hanh
ta c6
BA +
BC
+

OB
=
BD
+
OB
= Oli
+
BD
=
OD
Hinh
1.47
c) Theo cau b) ta c6 BA + BC
+
OB
=
OD
Theo quy t3c
trir
ta c6 MO -
MB
= BO
Ma
ob = BO suy ra BA
+
BC
+
OB
= MO
- MB

Bai
1.21: Cho tam giac ABC. Goi M, N, P Ian
lugt
la
trung diem ciia
BC,
CA, AB .
Chung minh rang
a)
NA +
PB
+ MC =
0
b) MC +
BP
+
NC =
BC
Huang dan gidi ^
a) Vi PB
= AP,
MC =
PN
nen
NA
+
PB
+ MC
-
NA + AP + PN

=
NP
+
PN
=
0
b)
Vi MC = BM va ket hgp voi quy tac ba
diem,
quy
tac
hinh binh hanh
ta c6
MC
+
BP
+ N(:
=
BM +
BP
+
NC:
=
BN + NC = BC u ,
Hinh
1.48
Bai
1.22: Cho hai
hinh binh hanh ABCD
va

AB'C'D'
c6
chung
dinh
A. Chung
minh
rang
B'B
+
CC' +
D'D
= d
Huang dan gidi , .
Theo quy tac
trir
va quy tac
hinh binh hanh
ta
CO
^ , ,.<^
B'B
+
CC' +
ITD
=
(AB -
AB')
+
(AC'
-

AC)
+
(AD
-
AD'
j^pl
=(AB
+AD)-AC-(AB'
+AD') +AC = 0
Bai
1.23: Cho ngu giac deu
ABCDE
tam O. '"^
Chung minh rang
OA
+
OB
+
OC
+
OE
+
OF
=
0
Huang dan gidi
Dat
u
= OA
+

OB
+ OC
+
OE
+
OF *
Vi
ngu giac deu nen vecto OA
+
OB + OC
+
OE
cung phuong
voi OF nen
u
cung phuong voi
OF .
•
. / <
Tuong
tu u
cung phuong voi
OE suy ra u = 0. ^ •
Bai
1.24: Cho
hinh binh hanh ABCD
.
Dung
AM
=

BA, MN
=
DA,
NP = DC,
PQ
=
BC. •••
Chung minh rang:
AQ =0 . A;
Humg dan gidi * j
'
Theo quy tac ba
diem
ta c6 AQ
=
AM
+
MN
+ NP +
PQ
=
BA
+ DA + DC
+
BC
Mat
khac BA + BC =
BD, DA
+ DC = DB suy ra AQ = BD + DB = 0
§3

TiCH CUA MOT VECTO VOI MOT SO
A. TOM TAT LY
THUYET
1.
Dinh nghia: Tich ciia
vecta
a vai so thuc k ^ 0 la mpt
vecto,
ki hieu la ka.
Cling
huong
voi a neu k > 0, ngUQic
huong
vai a neu k < 0 va c6 dp dai
bang
|k|
a •
Quy uoc: Oa = 0 va kO = 0
2. Tinhchat:
i)
(k + m)a = ka + ma ii) k(a ± b) = ka ± kb
iii)
k(ma) = (km)a iv) ka = Oc:> _ ^
^
,
v) la = a, (-l)a = -a
3. Dieu
kif
n
de hai

vecta
cung phirong
• b cung
phuong
a (a
5^
0) khi va chi khi
CO
so k
thoa
b = ka
• Dieu ki^n can va dii de A, B, C thang hang la c6 so' k sao cho AB = kAC
4.
Phan
tich mpt
vecto
theo
hai
vecto
khong cung phuong.
Cho a khong
cung
phuong b. Voi moi
vecta
x luon
duoc
bieu dien
X
= ma + nb vai m, n la cac so thuc duy nhat.
B. CAC

DANG
TOAN
VA
PHl/QNG
PHAP
GIAI.
DANG
TOAN
1:
DUNG VA TINHDO DAI VECTO CHUA TICH
MOT
VECTO VOI MOT SO.
1. PHUONG PHAPGlAl.
Su dung djnh nghia tich cua mpt
vecta
vai mpt so' va cac quy tac ve phep toan
•
vecta
de dung
vecto
chua
tich mpt
vecto
vai mpt s6^ ke't hap vai cac djnh li
pitago va he thuc lupng trong tam
giac
vuong de
tinh
dp dai ciia chiing.
2. CAC Vi DU.

Vi du 1: Cho tam
giac
deu ABC
canh
a . diem M la trung diem BC. Dung
cac
vecta
sau va
tinh
dp dai ciia chung.
a)lcB
+ MA b)BA-lBC
c)-AB + 2AC c)-MA-2,5MB
. 4
Laigidi
(Hinh 1.14)
) Do icB - CM suy ra
theo
quy tSc ba diem ta c6
18 . ^ '
1CB + MA = CM + MA = CA
2
Vay
b) Vi -BC = BM nen
theo
quy tac
trir
ta
1^
ICB

+ MA
2
= CA = a
CO
BA-^BC = BA-BM = MA
Theo
dinh li
Pitago
ta c6
MA
= 7AB2-BM2 - -'^
1
v2>
.^/3
Vay
1
BA—BC
2
= MA =
c) Gpi N la trung diem AB, Q la diem doi xung cua A qua C va P la dinh
cua hinh binh hanh AQPN.
Khi
do ta
CO
i
AB = AN, 2AC = AQ suy ra
theo
quy tac hinh binh hanh ta
CO
i

AB + 2AC = AN + AQ = AP
Gpi
L la hinh chie'u ciia A len PN
Vi
MN//AC => ANL = MNB = CAB = 60°
Xet tam
giac
vuong ANL ta c6
AL
- —S A I = A l\ Qin
AIMI
. =
2
a
2
sin
ANL
=
cos
ANL
=
AN
NL
AN
AL
= AN.sinANL =
-sin60°
= —
NL
=

AN.
cos
ANL
= - cos 60° = ^
3 93
Ta lai
CO
AQ = PN^PL = PN + NL = AQ + NL = 2a + - = —
Ap dung djnh li
Pitago
trong tam
giac
ALP ta c6
ax/2T
Vay
-AB+2AC
2
16 16
d) Gpi K la diem nkm tren doan AM sao cho MK = -MA, H thupc tia MB
sao cho MH =
2,5MB.
19
Khi
do -MA = MK,
2,5MB
= MH
3 =
Dodo-MA-2,5MB =
MK-MH
= Hk ,

4
Tac6MK
= ^AM = ^.^ = ^, MH =
2,5MB
= 2,5.^ = ^
4 4 2 8 2 4
Ap
dung dinh h'
Pitago
cho tarn tarn
giac
vuong KMH ta c6 ,;
KH
= VMH^ MK^ = . ^ = ^
16 64 8
Vay
-MA-2,5MB
4
= KH
a
VI27
8
\idu 2: Cho hinh vuong ABCD
canh
a . ., ' /
a) Chung minh r^ng u = 4MA - 3MB + MC - 2Ml5 khong phu thuoc vao vj
tri
diem M.
b) Tinh do dai
vecto

u .
Lm^»d«
(Hinh 1.15)
a) Goi O la tarn hinh vuong.
Theo
quy tac ba diem ta c6
u
= 4(Md + OA) - 3(Md + OB) + (MO + OC) - 2{Md + OD)
-40A-30B
+
OC-20D
Ma OD = -OB, OC = -OA nen u = 30A - OB
Suy ra u khong phu thuoc vao vj tri diem M
b) Lay diem A" trcn tia OA sao cho OA' = 30A khi do
OA'
= 30A do do u = OA' - OB = BA *
Mat
khac
BA' = VOB^ +OA'^ = VoB^
+90A^
=aS
Suy ra
|u|
= a>/5 '"^^^^ " '
3. BAI TAP
LUY^N
TAP.
Bai 1.25. Cho tam
giac
deu ABC

canh
a . Goi diem M, N Ian luot la trung
diem BC, CA. Dung cac
vecto
sau va
tinh
do dai ciia chung.
Ilinh
1.15
a) AN + ^CB
b)-BC-2MN c) AB + 2AC
Huang
dan
gidi
c) 0,2.5MA MB
2 .
a)
Theo
quy tac ba diem ta c6:
AN
+ -CB -
NC
+
CM
=
NM
2
Suy ra
AN
+ ^CB

= MN = -AB = -
2 2
1 . A'
b)
Theo
quy tac
trir
ta c6 -BC - 2MN = BM - BA = AM
1BC-2MN
2
= AM =
c) Goi F la diem do'i xirng ciia A qua C,
diem E la dinh ciia hinh binh hanh
ABEF
,
theo
quy tSc hinh binh hanh ta
CO AB + 2AC = AB + AF = AE
Goi I la hinh chieu ciia E len AC .
Yi AB / /EF ^ IFE = CAB = 60"
sin
IFE
= — ^
IE
= EF.sin
IFE
= a sin 60" = ^
EF
cos IFE = — => IF = EF. cos IFE =
a

cos 60" = -
EF 2
Ap
dung dinh li
Pitago
ta c6
I
AE =
N/AI^TIE^
=,
2a-f^
faTsf a^^
Suy ra
AB+2AC
= AE =
2)
3-^
d) Lay cac diem H,K sao cho
0,25MA
= MFI; -MB = MK
Suy ra
0,25MA
-1 MB = MH - MK = KH
Dodo
0,25MA-^MB| =
KH
=
,
AM
A2

r3
'-MB
2
8
3a
l4j
a>/39
8
II
4 )
Bai 1.26: Cho hinh vuong ABCD
canh
a .
a) Chung minh r3ng u = MA - 2MB + 3MC - 2MD khong phu thuoc vao vj tri
diem M.
b) Tinh do dai
vecto
u
Huong
dan
gidi
|
Goi O la tam hinh vuong.
Theo
quy tac ba diem ta c6
> '* 21
li
=
(MO
+

OA")
- 2
(MO
+
OB)
+ 3{M6 +
OC)
- 2
(MO
+ OD)
=
OA-20B + 30C-20D
Ma
OD
=
-OB,
OC
=
-OA nen u
=
-20A
k Suy ra u khong phu thu(
b)
li = -20A =20A
=
aN/2
Suy ra u khong phu thugc vao vj tri diem M 1
DANG
TOAN
2:

CHUNG
MINHDANG
THUC
VECTO.
1.
PHUONG
PHAPGIAI.
•
Si'e
dun^ cdc
kie'n
Ihiec sau
debich
ddi
venay
thanh
vckia
hoqc
cd hai
bieu
thirc
a
hai
vecung
blin<f
bieu
thiec
thu ba
hoqc
bicn

dot
tuan^ duang
vcdang
thi'tc
dung:
•
Cac
tinh
chat
phep
toan vecta
•
Cac quy t3c: quy tac ba diem, quy tac hinh binh hanh va quy tac
phep
tru \
•
Tinh
chat
trung diem: j
M
la trung diem doan thang
AB
o
MA
+
MB
=
0
M
la trung diem doan th5ng

AB
o OA
+
OB
=
20M
(Vai
O la diem tuy y)
•
Tinh
chat
trong tam:
G
la trong tarn ciia tam
giac
ABC
<=>
GA
+
GB
+
GC
=
O
G
la trong tam aia tam
giac
ABC
<=>
OA

+dB+OC
=
30G
(Voi
O la diem tuy y)
2.
CAC
VI Dg.
Vidu
1:
Cho tu
giac
ABCD
.
Goi
I,
J Ian lugt la trung diem ciia AB va CD, O
la
trung diem ciia IJ .Chung minh rang:
a)
AC
+
BD
=
2Ij
b)
OA
+
OB
+

OC
+
OD
=
d
* ^
c)
MA
+
MB
+
MC
+
MP
=
AMO
voi M la diem bat
ki.
'
Laigidi
(Hinh
1.16)
a) Theo quy tac ba diem ta c6
AC
=
Ai
+
lj
+
JC

,
Tuongtu
BD
=
Bi
+
r|
+
jD
Ma
I,
J Ian luot la trung diem cua
AB
va CD
nen
A"i
+
Bi
=
d,
JC
+
JD
=
0 ^
Vay
AC
+
BD
=

(Ai
+
Bl)
+
( JC
+
fd)
+
2IJ
=
2lj dpcm
b)
Theo he thuc trung diem ta c6 OA
+
OB
=
20i, OC
+
OD
=
20]
Mat
khac
O la trung diem IJ nen
OI
+
OJ
=
0
Suyra

OA+
OB+
OC
+OD
=
2(Oi
+
Oj)
=
6 dpcm
c)
Theo cau b ta c6
OA
+
OB
+
OC
+
OD
=
0
do do vai moi diem M thi
OA
+
OB
+
OC
+
Ob
=

d
o
(OM
+
MA)
+
(OM
+
MB)
+
(OM
+
MC)
+
(OM
+
MD)
=
0
'''
<j>
MA
+
MB
+
MC
+
MP
=
4MO

dpcm
Vi
du 2: Cho hai tam
giac
ABC va
AiB^Ci
c6 ciing trong tam G. Goi
Gi,
G2,
G3
Ian lugt la trong tam tam
giac
BCA^,
ABCj, ACBj.
Chung
minh
rang GG^
+
GG2
+
GG3
=
0
•
Lai
gidi
Vi
Gi la trong tam tam
giac
BCAj

nen
3GG,'
=
GB
+
GC
+
GAj
Tuong tu
G2,
G3
Ian lugt la trong tam tam
giac
ABC|,
ACBj
suy ra'
3GG2
= GA + GB + GC,' va 3GG3 = GA + GC + GB^
Cong
theo
ve'voi
vecac
dang
thuc tren ta c6
3(GGI
+ GG^ + GG3) = 2 (GA + GB + GC) + (GA^ + GBJ + Gq
M
Mat
khac
hai tam

giac
ABC va
A,BiC,
c6 cimg trong tam G nen
GA
+
GB
+
GC
=
0 va
GA^+
GBi+GCi
=
6
Suy ra GGi
+
GG2
+
GG£=d
VI
du
3:
Cho tam
giac
ABC c6 true tam H, trong tam G va tam
duang
tron
ngoai tiep O. Chung minh rSng: _ _
a)HA

+
HB
+
HC
=
2Hd b) OA
+
dB
+
OC
=
OH
c)
GH
+
2GO
=
0
—
Lmgifli
(Hinh
1.17)
a) Pe thay HA
+
HB
+
HC
=
2Hd neu tam
giac

ABC vuong
Neu tam
giac
ABC
khong vuong goi P
la
diem doi xung ciia A qua O khi do j
BH//PC
(vi
cung vuong goc vai AC)
BP//CH
(vi
cung vuong goc
voi
AB)
'
Suy ra BPCH la hinh binh hanh,
do do
theo
quy tic hinh binh hanh thi ^
HB
+
HC
=
HP (1)
Mat
khac
vi
O la trung diem ciia AD <
nenHA

+
HD
=
2Hd (2)
Tu
(1)
va (2) suy ra HA
+
HB
+
HC = 2H6
Minh
1.17
23
b)
Theo
cau
a) ta c6
HA+HB+HC=2HO
i,
o (HO +
OA) +
(HO + OB)
+
(HO + Oc)
=
2HO
<=>
OA + OB + OC = OH dpcm
c) Vi

G la
trong tam tam
giac
ABC
nen
OA
+
OB + OC =
30G
,
,
Mat
khac
theo
cau
b) ta c6
OA + OB + OC = OH
- Suy
ra
OH =
30G
c:> (OG +
GH)
-
30G = 0
«
GH + 2Gd =
0
Vi
du 4: Cho

tam
giac
ABC voi
AB =
c,
BC =
a,
CA =
b va c6
trong tam
G.
Goi
D,E,F
Ian lugt
la
hinh chieii
G
len
canh
BC,CA,
AB .
^
rlftif.
i
;|
Chung minh
rSng
a^ .GD +
b^
.GE + c^ .GF =

0
.
^
% LmgitJi (hinh
1.18)
Tren
tia GD, GE, MP Ian
luot
lay cac
diem
N, P, Q sao cho
GN
=
a,
GP = b, GQ =
c
va dung hinh binh hanh
GPRN
Ta CO a^GD + b^GE + c^GF
= 0
<=> a.GD.GN
+
b.GE.GP
+
c.GF.GQ
= 0 (*) >
Ta
CO
a.GD
=

2S\(3gc,
b.GE =
2S ,
C.GF
=
2S^GAB'
"^9*^
khac
G la
trong
tam tam
giac
ABC
nen
S,,GBC
S^^GCA
=
SACAB suy
ra
a.GD
=b.GE
=c.GF
Vay
(*)
GN + GP + GQ
= 0
Ta CO AC
=
GP = b,
PR =

BC = a
^
va ACB = GPR (goc
c6
cap
canh
vuong goc voi
nhau)
fj
Suy
ra
AACB =
AGPR(c.g.c)
/ ^
=> GR = AB = c
va
POR = BAC
Ta
CO
QGP + BAC =
]80"^QGP
+ GPR = 180°
=> Q,
G, R
thang hang do do
G la
trung diem ciia
QR
Theo
quy tMc hinh binh hanh

va
h^ thuc trung diem
ta c6 yy
GN
+ GP + GQ = GR + GQ = d
;„
Vay
a^GEJ
+ b^GE + c^ .GF =
0
.
•
Vi
du 5: Cho tam
giac
ABC voi cac
canh
AB =
c,
BC =
a,
CA
= b . Goi I la
tam duong tron noi tiep tam
giac
ABC.
•»* •
Chung minh rang alA + bIB + cIC =
0.
Lai

gidi
Cdch
1: (Hinh
1.19)
Goi
D la
chan
duong phan
giac
goc
A
24
Do
D la
chan
duong phan
giac giac
trong
gcSc
A nen
ta c6
^ =
^=>Bi3
= ^DC
A
DC
b b
ID
-
IB = ^(IC

-
ID) (b
+
c)ID = bIB + ciC
(1)
Do
I la
chan
duong phan
giac
nen
ta c6 :
ID
BD _ CD _
BD
+ CD a ^, t.J/i^3
D
Hinh
1.19 ^'
IA
BA CA
BA + CA b + c
i|
=:>(b + c)ID = -aIA
(2)
Tu
(1) va
(2) ta
CO
dieu phai chung minh

_
Cdch
2:
(hinh
1.20)Qua
C
dung duong thang
song song
vai
Al cat
BI tai
B';song
song
voi BI
cat
AI tai A'
d
Ta CO IC =
IA"'
+ IB'
(*)
Theo
djnh
ly
Talet
va
tinh
chat
duong phan
giac

trong
ta c6 :
IB^BA.=£=,iB
= -biB(l)
IB'
CA, b c
^ ^
Tuongtu: IA' = rA
(2)
c
Tu
(1)
va
(2) thay vao
(*) ta co :
IC =
-
-IA
- -
IB <» alA + bIB + cIC =
0
c c
3.
BAI TAP LUYgN TAP.
Bai 1.27:
Cho tam
giac
ABC. Goi M, N, P Ian
lugt
la

trung diem ciia
BC, CA, AB
.
Chung minh rang
* a
a) AM + BN + CP = d
b) OA + OB + OC = OKi + ON + OP voi
O
la diem bat ky.
Huang
dan
gidi
ui
a) AM + BN + Cr'=
j(AB.AC).l
b) OM
+
ON
+
OP
=
-^(0B + 0(:)
+
i(0C + 0A)
+
l(0A + 0B)
^
= 6A + OB + OC
Bai 1.28: Cho tam
giac

ABC. Goi
H la
diem doi xung voi
B
qua
G
vai
G la
trong
tam tam
giac.
Chung minh rang
' -
fei™
srfal
j
=
1(
AB
+ AC) + i(BC
+
BA) +
1(CA
+ CB)
=
6
a) AH = -AC-^AB, CH = AB AC
b)
MH=-AC AB
voi Mia trung diem cua BC

6 6 .'I; J
•
fM <•
Huang
dangidi
'' .• . .1 '.iCI
a) Ta
CO
AH = 2AG - AB = -(AC +
AB)
- AB = -
AC
- -
AB
i Of
1
/\:.)
Al
CH
= AH-AC = —AB —
AC
v:.,; -'Aif =• CHf:, u d)
3 3
b) MH =
-(AH-AB
+ CH] = -AC AB V")
2V / 6 6
Bai 1.29: Cho tam
giac
ABC c6 diem M thuoc

canh
BC.
Chiing
minh rMng AM = AB + AC
° ^ BC BC
Huang
dan
gidi
BC BC BC^ / BC\
MC-
MB;
= AM +—MB + —MC = AM
BC BC
Bai 1.30: Cho hai hinh binh hanh ABCD va AB'C'D' c6 chung dinh A. Chung
minh
r^ng B'B + CC' + D'D = 0
Huang
dan
gidi
Ta c6:
B^ + CC
+ =
( AB - AB') + (AC - AC) + (AD - AD')
=
(AB+AD)-AC-(AB' + AD')+AC = O
Bai 1.31: Cho n
vecto
doi mot
khac
phuong va tong ciia n -1

vecto
ba't ki
trong n
vecto
tren ciing phuong voi
vecto
con lai. Chung minh rang tong n
vecto
cho d tren bang
vecto
0
Huang
dan
gidi
Gia sir n
vecto
la aj, i = l,2, ,n . Dat u =aj + + a„
Vi
tong ciia n -1
vecto
bat ki trong n
vecto
tren ciing phuong voi
vecto
con
lai
do do u
Cling
phuong voi hai
vecto

aj, aj nen u = 0.
Bai 1.32: Cho tam
giac
ABC voi cac
canh
AB = c, BC = a, CA = b . Goi I la tam
va D, E, F Ian lugt la tiep diem ciia
canh
BC, CA, AB ciia duong tron noi
tiep tam
giac
ABC. M, N, P Ian lugt la trung diem ciia BC, CA, AB. Chung
minh
rang:
a)
B C)
cot—+
cot —
2 2
IA
+
f
C
cot — + cot —
2 2 )
IB +
A
B^
cot—
+ cot —

2 2J
ic = d
cotyIM
+ cot|lN + cot|lP = d
c) (b + c - a)IM + (a + c - b)IN + (a + b - c)IP = 0
d) aAD + bBE + cCF = d
• '
Huang
dan
gidi
a) Ggi r la ban
kinh
duong tron ngi tiep AABC ta c6
a = r
B C
cot —+
cot—
2 2
; b = r
cot—+
cot—
2 2
; c = r
J
Theo
vi du 5 ta c6 alA + bIB + cIC = 0
f
B C^
.
( c

A^
IB +
r
A B^
cot—+
cot—
IA
+
cot—
+ cot —
IB +
cot—
+
cot—
I
2 2,
I
2
2 J
I
2 2)
A
B>
cot—
+
cot—
2 2
IC
= 0
b) Ta

CO
IM = ^(iB + IC) ; IN = l(lC +
IA)
; IP = ^(lA + IB)
Theo
cau a) ta c6 coty(iB + ic) + cot|(lA + ic) + coty(lA +
IB)
= 6
Suy ra cot
—IM
+ cot-IN +
cot-IP
= 0
^222
c) Ta
CO
D M
IA
= IP + IN -
IM;
IB = IM + IP -
IN;
IC = IM + IN - IP
A
Ket hgp vi du 5 suy ra
alA + bIB + cIC = 0 . f
<i>(b + c-a)lM + (a + c-b)iN + (a +
b-c)iP
= 6 P
d)rD =

^IB 5^IC
= iP^IB.iPl^IC
BC BC a a
oaID = (p-c)lB + (p-a)lC voi p la nua chu vi.
Hinh
I.5I
Tuong tu ta
CO
: i{A
biE = (p-a)lC + (p-c)lA;cIF = (p-b)lA + (p-a)lB
=>alb + bIE + cIF = (2p-b-c)lA +
(2p-c-a)lB
+
(2p-a-b)lC
= alA + bIB + cIC
=>
a
AD
+ bBE + cCF = 0
Bai 1.33: Cho tam
giac
ABC . M la diem bat ky n^m trong tam
giac.
Chung minh rang : SMBC MA+ SMCAMB+ SMABMC=d
Huang dan
gidi
A'
C A'
B
Goi

A'
la
giao
diem
AM
v6i
BC
ta c6
MA'
=
MB
+
_
MC
(*)
Mat
khac:
— - ^MK^ -
^MAC
A'B
S
MAB
'MAB
A'C
A'B
+ 1
=
BC
S
BC

_fMAC
'MAB
A'B
^MAB
BC
^MAB
+^MAC
Va
A'C
^MAC
Va
BC
^MAB
+
^MAC
0)
Mat
khac
MA' = -^^MA
=
•
i-Mj
ion HOT!
'MBC
^MAB
+
^MAC
MA
(2)
MA

Thay
(1)
va
(2)
vao (*) ta dirge dieu phai chirng
minh.
Bai
1.34:
Cho
da
giac
loi A,
Aj A^
(n >
3);
ejj
< i < n
la
vecto
dan
vi
vuong
gocvai
AjAj^, (xem A„+| = A,)
va
huang ra phi'a ngoai da
giac.
Chung minh rMng:
A,A2e]
+

A2K^e2
+
-
+
AnA,e„
=
0
(dinh
ly
con nhim)
Huang dan
gidi
Ta chung minh bang quy nap
Vai
n
=
3
dang thuc tro thanh
a.e,
+b.e2
+c.e3
=0
(diing
vi
dang thuc
nay
tuang
duong vai dang thuc
a
bai

11)
Gia su dung voi n
=
k-l,
k>4
Goi
e la
vecto
don
vj
vuong
goc voi
A,A|^_,
va
huong
ra
ngoai
tam
giac
'
Theo
gia thiet quy nap ta
c6
A,A2e,+A2A3e2+
+
A,„2A,_,e,:2+A^_,A,(-e)
= d
(1)
MatkhacxettamgiacA,A|^_,A,^
taco

>
A]Ak_,e
+ A^_,Ake^_; + A^A,e^=0 (2)
Tir
(1)
va
(2)
suy ra
dieu phai chirng
minh.
Bai
1.35:
Cho
da
giac
loi
A]
A2
An
(n
>
3) voi
I la
tam duong
tron
tiep xiic
cac
canh ciia
da
giac;

goi ej,l < i < n
la vec to
don
vi
ciing huong voi
vec to
A,
lAj.
Chung minh rang
cos—i-e,
+
cos-^e,
+ +
cos—^e
"
=
0
Huang dan
gidi
Goi Bj, i =
1,2, ,n
la cac tiep diem duang
tron
noi tiep vai canh A^Aj^j
Xettugiac A,B,IB„
c6
A^i;,!
=
A^B,!
=90"

va
B;A,I
= B7A,I
Suy ra
B^, =
B^IA,
.
Mat
khac
IB.
=
IB„ do do lA,
±
B,B„
Tuong tu ta c6 lA;
±
Bj„,Bj,i =
2,3, ,n
Xet
da
giac
loi
B,B2 Bn
theo dinh
ly
con nhim ta
c6
B„ B, .e; + B, B2 .e2 + + B„ _, B„ .e„ =
6
Hiiih

1.54
A
• A • A - -
o
IB|.cos-y-e,
+
IB2.cos-Y-e2+•••
+I^n'^"^"^'-'n
Ma IB,
=
IB2 = = IB,,
suy
ra dpcm.
Bai
1.36:
Cho tam
giac
ABC
vuong tai
A. I la
trung diem
cua
duang
cao
AH.
Chiing
minh rang
:
a^
lA

+
b^
IB + c^ IC =
0 .
Huang dan
gidi
^
. HB
HB.BC
c^
Ta CO
HC
HC.BC
'
b^
• c
Suy ra
111=-^
:rIB +
c^+b^
c^ + b^
IC
b^
„ c^
Ma b^ + c^ = a^
va
IH = -lA nen
suy
ra -IA =
—

IB
+ — IC
a
a
Hay aMA + b^IB + cMC =
0.
DANG TOAN
3:
XkC DINH
DIEM
M THOA MAN MOT
DANG
THUC VECTO CHO TRUOC
1.
PHUONG PHAPGIAI.
•
Ta
bie'n doi d5ng thuc
vecto
vc
dang AM = a trong do diem
A va a da
biet.
Khi
do
ton tai duy nha't diem
M
sao cho AM =
a, de
dung diem

M ta lay A
lam
goc
dung mot
vecto
b5ng
vecto
a suy
ra diem ngon
vecto
nay chinh
la
diemM.
•
j^^^
-
•
Ta
bieh doi
ve
d3ng thuc
vecto
da
biet ciia trung diem doan thang
va
trong
tam
tam
giac.
1^

29
2. CAC VJ
Dg.
Vidu
1: Cho hai
diem
A, B
phan biet.
Xac
djnh diem
M
bie't 2MA
-
3MB
= 0
LaigidiiKmh
1.21)
.
Taco
2MA-3MB
= ()
«2MA-3(MA
+
AB)
= d ^ ^
oAM
=
3AB
M
nam tren

tia AB va
AM
= 3AB - i i'
Vidu
2: Cho tu
giac
ABCD
. Xac
djnh diem M,N,P
sao cho
a) 2MA + MB + MC =
0 b)
NA
+
NB + NC + ND
= 0
c)
3PA
+ PB
+
PC
+
PD
= 6
LOT^ifli(hlnh
1.22)
a) Goi
I la
trung diem
BC suy ra

MB
+
MC
=
2Mi
Dodo 2MA
+
MB
+
MC
= 0
« 2MA
+
2Mi
=
d
c>
MA + MI
= 0
Suy
ra M la
trung diem AI
b)
Goi K, H Ian
luat
la
trung diem ciia AB,
CD ta c6
NA
+ NB + NC + ND

=
d
<=>
2NK + 2NH
= 0
« NK
+
NH
= d
<r>
N la
trung diem
cua
KH
c) Goi G la
trong tam tam
giac
BCD
khi
do ta c6
PB +
PC
+ PD
= 3PG
Suy
ra 3PA + re + PC + PD =
d<=>3PA
+ 3PG = 0
Hinh
1.22

<=:>PA
+ PG =
6<:z>P
la
trung diem AG
.
Vi
du 3: Cho
truoc
hai
diem
A, B va hai so
thuc
a, P
thoa
man a + p ^ 0.
Chung minh rang ton
tai
duy nha't diem
I
thoa
man
alA
+
piB
= 0.
Tu
do,
suy ra
vai diem

bat
ki
M
thi aMA
+
PMB
= (a +
p)MI.
Lai
gidi
Ta
c6:
alA + pre
=
d
<:> alA + P(IA + AB)
=
d
0
(a
+ p)IA + pAB
=
d.
«(a
+ P)Ai
=
pAB
o
AI
= BA.

a + p
Vi
A, B
CO
djnh
nen
vecto
—^BA
khong doi,
do do ton tai
duv nha't diem
a + p
• •'
1
thoa
man dieu ki^n.
Tu
do
suy ra aMA + pMB = a(MI + lA) + P(Mi +
IB)
v,;.
=(a +
p)MI
+
(aIA
+
pIB)
=(a +
p)Mi dpcm.
3n

B
M
Hinh
1.21
3. BA! TAP LUVeN TAP.
Bai
1.37: Xac
djnh
cac
diem I,
J, K, L
bie't
3)IA-2rB = 6^ b)JA-JB-2JC = 6 ^
c)KA
+ KB + KC = BC d)
2LA-LB
+3LC
= AB +
AC
Huong
dan
gidi
a)
I la
diem doi xung
cua A qua B. /•}
rr^lAB
c)Ak
= -AB
d)AL

= -BC ^ ^ ,
Bai
1.38: Cho tu
giac
ABCD. Tim diem
co
dinh
I va
hang
so k de he
thuc
sau
thoa
man voi moi
M
a)MA + MB + 2MC
=
kMI
b)
2MA + 3MB
-
MD
=
kMI
c)
MA + 2MB + 3MC
-
4MD
=
kMI

Huang
dan
gidi
a)
Cho
M^I^iA
+ iB + 2fc =
0oli
+
IC:
= 0
Vol
J la
trung diem
cua
AB,
suy ra I la
trung diem ciia
JC
,
>(
x
MA
+
MB
+
2MC
=
kMi<=>4MI
=

kMi=>k
= 4 ^
b)
k = 4,
Ai
=
-(3AB-AD)
^
c) k = 2,
IA
=
2AB
+
3AC-4AD
Bai
1.39: Cho tam
giac
ABC va ba so
thuc
a, p, y
khong dong thai
bang
khong.
Chung minh rang:
a)
Neu a
+ p + y
^ 0
thi
ton tai

duy nha't diem
M sao cho
aMA
+
pMB + yMC
=
d.
b)
Neu a
+ P + y
= 0
thi khong ton
tai
diem
N sao cho
aNA
+
PNB
+
yNC
= d.
Huang
dan
gidi
a)
Vi a
+ p+y#0=^(a + p)
+ (p
+ Y)
+

(Y
+ a)*0
Khong
mat
tinh
tong quat
gia su a + P ^ 0
=> 3!D: aDA
+
pDB
= 6. I
Suy
ra
aMA
+ pNS +
yMC
=
0<::>(a
+
p)MD
+
yMC
= 0 '
Do
do
ton
tai
duy nha't diem
M
b)

Gia sir
ton
tai
diem
N va a ^ 0 '/^ .7
Taco
aNA +
pNB+yNC=6<:>CA
=
—CB
(mau thuan voi
ABC la
tam
giac)
mmrn^.
31
Bai
1.40: Cho n diem A|,A2, ,
A„ va n
scV k,,
k2, ,
k„
ma
k,
+ k2 +
•••
+
k„ = k 0 , , . ,
a) Chung minh rang c6 duy nhat diem
G

sao cho
k,GA,
+
kjGAj
+

+
k„GA„
=
0
.
Diem
G
nhu
the goi la
taiti
ti cir
cua
Iw
diciu
Aj
\^(in
vai he so'
k,
.
Trong
truong
hop cac
he so k;
bang nhau(ta

c6
the chon cackj deu bang 1
)
thi
G
goi
la
fr('//y
taifi ci'in
he
diem
A;
b) Chiing minh rang neu
G la
tam ti cu noi
o
cau
a)
thi voi diem
M
bat
ky
ta
c6
-(k,MAi
+
k2MA2
+

+

k„MA„")
=
OG
Ihiang
dan
giiii
t-A v:;,m
ur^ nun
Kr.ii-
O
la diem
tu\
v,
ta c6:
if
k,GA,
+
k.GA,
+ +
k„GA,,
=0
k, (OA,
-
OG)
+ k2 (OA2 -
OG)
+ + k„
(OA,,' -OG)
= 0
o OG =

^(k,OA,
+
k20A2
+ +
k„OA„')
Suy ra
G
xac
dinh
duy nhat.
DANG
TOAN
4:
PHAN
TICH
MOT
VECTO THEO
HAI
VECTO
KHONG
CUNG
PHUONG.
1. PHUONG PHAPGIAI.
Sir dung
cac
tinh
chat phep toan
vecto,
ba
quy tac phep toan

vecto
va
tinh
chat
trung
diem, trong tam trong tam
giac.
2.
CAC Vi Dg.
Vidu V. Cho tam
giac
ABC . Dat
a =
AB,
b = AC .
a) Hay dirng cac diem M,
N
thcSa man:
AM = -
AB,
CN =
2BC
3
b) Hay phan
tich
CM,
AN, MN
qua cac
vec to a va b .
c) Goi

1
la
diem thoa: MI
=
CM. Chung minh
1,A,N
thcing hang
.
LOT^ifli (hinh
1.23)
1
a)
Vi AM =
-AB suy
ra M
thuoc canh
AB va
AM
=
^AB;
CN =
2BC, suy ra
N
thucV tia
BCvaCN
=
2BC.
C Hinh
1.23
Cty

TNHH
MTV DWH
Khang
Viet
b)
Taco:
CM
=
CA +
AM
=-SC+
|AB
= |a-b
AN
= AB
+
BN = AB
+ 3BC
= AB
+
3(AC-AB)
=
-2a
+
3b
1
_ . -
7
-
MN = MA + AN

= a -
2a + 3b
= a
+ 3b.
r\:
M =
AM
+
MI
= iAB
+CM = ^a+ ia-b
=-^(-^^
*-/
3 3 3 3
^ AI
= - -
AN =>
A,
I,
N
th5ng hang.
3
Vi
du
2:
Cho tam
giac
ABC, tren canh
BC lay M sao
cho BM

=
3CM, tren
doan
AM
lay
N
sao cho 2AN
=
5MN .
G
la trong tam tam
giac
ABC
.
a) Phan
tich
cac vecta AM,
BN
qua cac
vec to AB va AC
b) Phan
tich
cac
vecto
GC,
MN
qua cac vec to
GA va GB
Lai
^idi (hinh 1.24)

3
_ 5 •
a)
Theo
gia thiet ta c6: BM =
-
BC
va AN = -
AM
suyra
AM
=
AB + BM
=
AB + |BC
= AB +-(AC
-
AB) =
i
AB + - AC
4V
/ 4 4
BN
= BA + AN =
-AB
+
-AM
7
Hinh
1.24

=-AB+-
7
fi
. 3 .
-AB+-AC
U
4
23
15
=
-—AB
+—AC
28
28
b) Vi
G
la trong tam tam
giac
ABC
nen
GA +
GB
+
GC
=
0 suy ra
GC=-GA-GB
Taco
^0^J
=

AM
=
7
7
-AB+-AC
.4
4
= _ J_(GB
-
GA)
-
—(GC
-
GA)
14V
/ 14V /
= _i-(GB
-
GA)
-
—(-GA
-
GB
- GA)
14V
/ 14V /
=-GA+-GB
2
7
VI

du 3: Cho
hinh
binh
hanh
ABCD
. Goi M,
N
Ian luot la hai diem n^m tren
hai
canh
AB va CD
sao cho AB
=
3AM, CD
=
2CN
va G la
trpng tamjtam
giac
MNB
.
Phan
tich
cac
vecto
AN, MN,
AG
qua cac
vec
to^ABj^a^C^

Lcrigidi{h\nh
1.25)
1
^
A
Ta c6: AN = AC + CN = AC - - AB
2
MN
= MA + AN = AB + AC - - AB
3
M
= AB+AC
6
Vi
G la trgng tam tam giacMNB nen '
' D
B
3AG = AM + AN + AB = - AB +
3
1
AC AB
2
Hlnh
1.25
5 -
+ AB = -AB + AC \
;Suy ra AG = —AB + -AC
18 3
3. BAI TAP LUYgN TAP.
Bai 1.41: Cho tam

giac
ABC .Lay cac diem M,N,P sao cho MB = 3MC,
NA
+ 2NC = 6, PA + PB = d
a)
Bieu
dien cac
vecto
AP, AN, AM
theo
cac
vecto
AB va AC
b)
Bieu
dien cac
vecto
MP, MN
theo
cac
vecto
AB va AC
Chung minh
rang
ba diem M, N, P th^ng
hang?
Huong
dan
gidi
a) AP = 1 AB, AN = -

AC,
AM = - AC - i AB
2 2 2 2
3
1
b) MP = AB AC,MN = -AB AC
2 2 4
MP = 2MN => M, N, P th5ng hang
Bai 1.42: Cho tam
giac
ABC.Goi I, J la hai diem xac djnh boi
iA
= 2IB, 3JA + 2JC = d
a) Tinh IJ
theo
AB va AC.
b) Chung minh duong th^ng IJ di qua trong tam G ciia tam
giac
ABC.
Hu&ng
dan
gidi
a) IJ = -2AB + -AC i
b) IG = AB+iAC=>5ij = 6IG suy ra IJ di qua trong tam G cua tam
giac
ABC.
Bai 1.43. Cho tam
giac
ABC c6 trong tam G. Goi I la diem tren
canh

BC sao
cho 2CI = 3BI va J la diem tren BC keo dai sao cho 5JB = 2JC .
a) Hay phan tich AI, AJ
theo
AB va AC .
34
b) Hay phan tich AG
theo
AI va AJ .
Huang
dan
gidi
3
27.
a) Ta CO
• 2IC =
-3IB<=>
AI = -AB + -AC. «
5 5
5j^ = 2JC « 5( AB - AJ) = 2( AC - AJ) « AJ = | AB -1 AC •
b) Gpi M la trung diem BC, ta co:
AG
=
I
AM
= ||(AB + AC) = 1( AB + AC) => A~G = ^
AI
-:! AJ.
Bai 1.44: Cho hai
vecto

a, b khong ciing phuong. Tim x sao cho ,M a
a) u = a +
(2x-l)b
va v = xa + b ciing phuong
b) u = 3a + xb va u=(l-x)a b cung huong
Huong
dan
gidi
a) u cimg phuong voi v <» c6 so
thuc
k sao cho
UK,,;':
u
= kv <=> a+
(2x-l)b
= k xa + b
kx = l
k
= 2x-l
x = l
b) u ciing huong vai v <=> c6
so'thuc
k duong sao cho
u
= kv => 3a + xb = k
3 = k(l-x)
3
k
= -
2 => x = -1

k
= -3(1)
DANG
TOAN
5: CHUNG MINH HAI DIEM TRUNG
NHAU,
HAI
TAM GIAC CUNG TRONG TAM
1.
PHUONG
PHAPGIAI.
• De chung minh hai diem Aj va A2
triing
nhau, ta lya chpn mpt trong hai
each
sau :
Cdch
1: Chung minh
A1A2
=0.
Cdch
2: Chung minh OAi = OAj voi O la diem tuy y.
De chung minh hai tam
giac
ABC va A'B'C cimg trong tam ta lam nhu sau:
Cdch
1: Chung minh G la trong tam AABC
triing
voi G' la trong tam
AA'B'C

Cdch
2: Goi G la trong tam AABC (tuc ta c6 GA + GB + GC = 0) ta di chung
minh
GA' + GB' + GC' = 0.
PJiaii loiii vc'i
phutriig
phiip
^it'ii IIluli hoc 10
2. CAC
VI
Dg. *
Vt du 1: Chung minh rang AB = CD khi va chi khi trung diem cua hai doan
thang AD va BC
trimg
nhau.
Goi
I,
J Ian lugt la trung diem cua AD va BC suy ra AI = ID, CJ = JB
Do do AB = CD » AI + IJ + JB = CJ + JI + ID
o ij = Jl o IJ = d hay I
triing
v6i J , , j, r
Vi du 2: Cho tam
giac
ABC, tren cac canh AB, BC, CA ta lay Ian lugt cac
NT
O , AM BN CP „ , , .
diem
M, N, P sao cho = = . Chung mmh rang hai tam
giac

AB BC CA ^ ^ ^
ABC va MNP c6 cung trgng tam. ^ " ^
Lai
gidi
AM
Gia su ^ = k suy ra AM = kAB ; BN = kBC ; CP = kCA
Cdch
1: Ggi G, G' Ian lugt la trgng tam AABC va AMNP
Suy ra GA + GB + GC = 0 va G^ + G>J + GT = 0 (*)
Ta CO AM = kAB « AG + GG' + GlA = kAB
Tuong tu BG + GG' +
GT^J
= kBC
Va CG + GG' + G'P = kCA
Cong ve v6i ve
tirng
dang thuc tren ta dugc
(AG
+ BG +
CG)
+
SGG"
+ (G'M + G'N + GT) = k(AB + BC + CA
Ket hgp vai (*) ta dugc GG' = 0 j
Suy ra dieu phai chung minh
Cdch
2: Ggi G la trgng tam tam
giac
ABC suy ra
GA

+ GB + GC = 6
Taco:
GM +GN + GP = GA + AM + GB + BN + GC +
CP
= AM + BN + CP = kAB + kBC + kCA =
k(
AB + BC + CA) = 0
Vay hai tam
giac
ABC va MNP c6 cung trgng tam.
Vidu
3: Cho luc
giac
ABCDEF. Ggi M, N, P, Q, R, S Ian lugt la trung diem
cua cac canh AB, BC, CD, DE, EF, FA. Chung minh rang hai tam
giac
MPR va NQS c6 cung trgng tam. "'^ ^ '
Lai
gidi
(hinh 1.26)
Ggi
G la trgng tam cua AMPR suy ra GM + GP + GR = 6 (*)
Matkhac 2GM = GA + GB, 2GP = GC + GD, 2GR = GE + GF.
=> 2(GM + GP + GR) = GA + GB + GC + GD + GE + GF
Ket hgp vai (*) ta dugc
GA + GB + GC + GD + GE + GF = 6
^ (GA + GF) + (GB + GC) + (GD + GE) = 6
A
E
Hinh

1.26
<=> 2GS + 2GN + 2GQ = 0
<:r>GS
+ GN + GQ = 0
Suy ra G la trgng tam ciia ASNQ . '^0^: •
Vay AMPR va ASNQ c6 cimg trgng tam.
Vidu
4: Cho hai hinh binh hanh ABCD va AB'C'D' chung
dinh
A. Chung
minh
rang hai tam
giac
BCD va B'CD' cung trgng tam.
Lm^ia;
(hinh 1.27)
Ggi
G la trgng tam tam
giac
BC' D ' , #
suy ra GB + GC' + GD - 0 B
oGB' + GC + GD' + WB + CC + D'D = 0 (1)
Mat
khac
theo
quy tac phep tru va hinh binh
hanh ta c6
B'B + CC' + D'D
= (AB -
AB^')

+ (AC' - AC) + (AD -
AD''
= (AB +
AD)
- AC - (AB' + AD') + AC
=
AC-AC-AC'
+ AC = 0 (2)
Tu
(1) va (2) ta
CO
GB" + GC + GD" = 0 hay G la trgng tam tam
giac
B' CD'
3. BAI TAP
LUY^N
TAP.
Bai 1.45. Cho cac tam
giac
ABC, A'B'C c6 G, G' Ian lugt la trgng tam. Chung
minh
rang:
AA'+
BB'+ CC = 3GG'. Tix do suy ra dieu ki^n can va dii de
hai
tam
giac
c6 cimg trgng tam .
Huang
dan

gidi
Taco
A>V + BB' + CC v
= (AG+GC+G^V)+(BG+GC+CB')+(CG^
= 3GC +
(AG
+ BG +
CG)
+
(C^ + G^ + CC') = 3GC '
Suy ra dieu kien can va du de hai tam
giac
c6 cimg trgng tam la
AA'
+
BB'^
+ CC' = 0
Bai 1.46. Cho tam
giac
ABC, ve cac hinh binh hanh ABIJ,
BCPQ,
CARS.
Chung minh rang ARIP,
AJQS
c6 ciing trgng tam. /
Huang dan gidi
G
la
trong
tam

ARIP =>
GR +
GI +
GP =
6
Ta
CO
RJ
+ IQ +
PS
= (RA
+
JA) +
(iB
+ BQ) +
(PC
+
CS)
=
RA
+
CS)
+
(fA + IB)
+
(PQ
+
PC)
=
0

Suy
ra
GR
+
a
+
GP
=
Q
+
GQ
+
GS
GJ
+
GQ
+
GS
= d
Do
do
G la
trong
tam AJQS
.
Bai
1.47.
Cho tu giac
ABCD
. Goi M,

N,
P, Q
Ian
lugt
la
trung diem ciia
AB,
BC,
CD, DA. Chung minh rang
hai tam giac ANP
va
CMQ
c6
cung trong
tam.
Huang dan gidi
G
la
trong tam
AANP =>
GA
+ GN +
GP =
6 ur,.)
1
-r^ 1
Taco AC +
NM +
PQ =
AC AC-iAC

=
0
Suy
ra
GA
+
GN
+ GP = GC
+ GM +
GQ^GC +
GM
+ GQ
=
d
Do
do
G la
trpng
tam
ACMQ
.
Bai
1.48. Cho tam
giac
ABC. Goi A', B' ,C' la cac
diem
xac
djnh
boi
2011A7B

+
2012A'C
=
0,
ZOllB^C +
2012B'A =
0 ,
2011C
A
+ 2012CB =
0
Chung minh rang AABC
va
AA'B'C cung trong
tam
Hucmg dan gidi
G
la
trong
tam
AABC =:> GA +
GB
+
GC =
0
Taco 20nA^
+
2012A7c
=
0

o 201I(A'A
+
AB) +
2012(A7^
+ AC) =
0
c:>4023A~A + 20nAB
+
2012AC =
0 (1)
Tuong
tu
ta
CO
.
4023B'B
+
201IBC
+
2012BA
- 0 ;
4023C'(:
+
20nCA
+
2012CB =
6
Cpng
ve
vai

ve'lai
ta
dugc
4023(
AA'
+
BB'
+
CC')
+ BA +
AC + CB
=
OoAA^
+
BB'
+
CC'
=
0
Suy
ra
GA
+
GB
+
GC
=
GA^' +
GB'
+ GC'

=>
GAT' +
GB'
+ GC"' =
0
Do
do
G la
trong
tam
AA'B'C.
Bai
1.49.
Cho
AABCva AA'B'Cc6 ciing trong
tam
G, ggi
Gi,G2,G3la trong
tam
cac tam
giac
BCA',CAB', ABC .Chung minh rang
AGiG2G3Cung
c6
trpng
tam G.
Huang dan gidi
Vi
AABC
va

AA'
B'C
CO
ciing trong
tam
G
suy
ra
AA"
+
BB' +
CC =
0
38
Vi
G|,G2,G3la trong
tam
cac
tam giac
BCA',CAB',
ABC
nen
3AGi
+
3BG2
+
3CG3
= (AB + AC + AA') + (BC +
BA
+

BB')
+
(CA +
CB
+
CC')
=
AA'' + BB''
+
CC'' =
6
Suy
ra
AG,
+
BG2
+
CG3
=0
do do
G la
trong
tam
AG,G2G3.
Bai
1.50. Cho tu
giac
ABCD
c6
trong

tam G. Goi
G],
G2,
G3,
G4
Ian
lugt
la
trgng
tam
cac
tam giac
AABC, ABCD, ACDA, ADAB. Chung minh
r^ng
G
cung
la
trgng
tam tu giac
G,G2G3G4.
Huang dan gidi
G
la
trgng
tam tu giac
GJG2G3G4
0
GGj
+
GGj +

GG3 + GG4
=
0 (*)
Vi
G, la
trong
tam
AABC =>
GA
+
GB
+
GC =
3GGi,
tuong
tu ta
c6
GB
+
GC + GD =
3GG2
,
GC +
GD + GA
=
3GG3
,
GD + GA + GB =
3GG4
Do

do
n «
3(GA
+ GB
+
GC
+
GD)
=
0
(dung) dpcm
Bai
1.51.
Cho tam giac
ABC
deu
va M la
mot
diem
nam
trong
tam giac. Goi
Ai,B|,Ci
Ian
lugt
la
diem
doi
xung
M qua BC,

CA, AB.
Chung minh rang
tam
giac ABC
va
A,B,C,
c6
ciing trgng
tam.
Huang dan gidi
Ggi
D, E,
F
tuong
ung
la
giao
diem ciia MA,,
MB,, MC, voi
cac
canh
BC,
CA, AB.
O
la
trgng
tam deu
AABC
Ta
CO

MA,"
+ MB,' + MC, =
2(MD +
ME +
MF'
rr;;
3;
Mat
khac theo bai tap
6
(dang
2)
thi
wb
+
ME + MF =
^MO
Suy
ra
MA, + MB,
+
MC,
=
3MO
do do
O la
trgng
tam tam giac
A,B,C,
Bai

1.52. Cho
tam giac ABC,
diem
O
nam
trong
tam
giac.
Ggi
A,,Bi,C,
Ian
lugt
la
hinh
chieu
ciia
O len BC,
CA, AB.
Lay cac
diem A2,B2,C2
Ian
lugt
thugc
cac tia
OA,, OB,, OC,
sao cho
OA2
=a,
OB2
= b,

OC2
=
c .
Chung
minh
O la
trgng
tam tam giac
A2B2C2.
Huang dan gidi
Ta
CO
OA2
+
OB2 +OC2
= ^ .
OB, OC,
=
+
b??^
+
=
0
(Theo
djnh
ly con nhim)
OA,
- +
c-
OBi

OC,
Do
do
O la
trgng
tam tam giac
A2B2C2
39
DANG
TOAN 6: T/M TAP HOP DIEM THOA MAN
DIEU
KIEN
VECTO CHO TRUOC.
1. PHLfONG PHAPGlAl.
De tim tap hap diem M
thoa
man man dieu ki?n
vecto
ta quy ve mot trong
cac dang sau:
- Ne'u MA = MB voi A, B phan biet cho tnroc thi M thuoc duong trung true
aia doan AB.
voi
A, B, C phan biet cho triroc thi M thuoc duong tron
- Neu
MC =k.
AB
tarn C, ban
kinh
bang

k. AB
- Ne'u MA = kBC voi A, B, C phan bi^t va k la so' thuc thay doi thi . ^
+ M thuoc duong thang qua A
song song
voi BC voi k e R
+ M thuoc nira duong thang qua A
song song
voi BC va ciing huong BC
voi
k > 0
+ M thuoc nua duong th^ng qua A
song song
voi BC va
ngu-oc
huong BC
vol
k<0
'•i
Ne'u MA = kBC, B^C voi A, B, C thang hang va k thay doi thi tap hop
diem M la duong thang BC.
2. CAC Vl DU.
Vidu
1: Cho tam
giac
ABC
a) Chung minh
rSng
ton tai duy nha't diem I
thoa
man : 2iA + 3IB + 4iC = 0.

b) Tim quy tich diem M
thoa
man : |2MA + 3MB +
4MC|
=
|MB
- MA
Liri
gidi
a)
Taco:
2IA + SIB + 4iC =
0<=>2iA
+ 3(IA +
i^)
+ 4(iA + AC) = 6
o 9IA = -SAB - 4AC <^ lA
3AB + 4ACI
>
I ton tai va duy nha't.
b) Voi I la diem
duoc
xac djnh 6 cau a, ta c6:
2MA + 3MB + 4MC = 9Mi + (2IA + 3IB + 4IC) = 9MI va MB - MA = AB nen
12MA + 3MB + 4MCIHMB - MA lol 9Mi W AB io MI = ^
Vay quy tich ciia M !a duong tron tarn I ban
kinh
AB
Vidu
2: Cho tam

giac
ABC. Tim tap hop cac diem M
thoa
man dieu kien sau:
a)
|MA
+
MB|
=
|MA
+ MCl
b) MA + MB = k(MA + 2MB - 3MC) voi k la so thuc thay doi
40
Lai
gidi
(hinh 1.28)
a) Goi E, F Ian luot la trung diem ciia AB, AC suy ra
SJA + MB = 2ME va MA + MC = 2MF
Khi
do IMA + MB = MA + MC
|2MF|<»ME = MF
2ME
Vay tap hop cac diem M la duong
trung
true
ctia
EF
b) Ta CO
MA
+ 2MB - 3MC = MA + 2(MA + AB) - 3(MA + AC)

= 2AB - 3AC = 2AB - 2AH = 2HB
Voi
H la diem
thoa
man AH = — AC

.I.:.:; .
^
Suy ra MA + MB = k(MA + 2MB - 3MC) » 2ME = 2kHB « ME = kHB
Vay tap hop diem M la duong
thSng
di qua E va
song song
voi HB
Vi du 3: Cho tu
giac
ABCD. Voi so k tuy y, lay cac diem M va N sao cho
AM
= kAB, DN = kDC. Tim tap hop cac trung diem I cua doan thing
MN
khi k thay doi.
LOT^ifli
(hinh 1.29)
Goi O, O' Ian lugt la trung diem ciia AD va BC, ta c6
AB = Ad + OO" + O'B va DC = DO + OO' + OC
Suy ra AB + DC = 200'
Tuong tu vi O, I Ian luot la trung diem ciia AD
va MN nen AM + DN = 20i
Dodo Oi = i(kAB + kDC) = kOO' 1
Hinh

1.29
Vay khi k thay doi, tap hop diem I la duong thang OO'
3.
BAI TAP LUYlN TAP.
Bai 1.53. Cho 2 diem co djnh A, B. Tim tap hop cac diem M sao cho:
a) IMA + MBI = IMA - MBI b) |2MA +
MB|
-
|MA
+ 2MB|
Huang
dan
gidi
3) Tap hop diem M la duong tron tam I ban
kinh
voi I la trung diem cua
AB.
b) Goi K la diem
thoa
man: 2KA +
KB
= 0
ao<.
L
la diem
thoa
man:
LB
+ 2LC = 0
41

Ta
c6:
2MA
+ MB
ML
MB + 2MC
«
MK
Tap hop diem
M la
duang trung true cua doan thang KL.
Bai 1.54. Cho AABC. Tim tap hop
cac
diem
M sao
cho:
a) MA + kMB = kMC voi
k la so
thuc thay doi
b)
V
= MA + MB + 2MC cung phirong vol
vec to BC
c)
IMA
+ BCl =
|MA
-
MB| (HD:
dung hinh binh hanh ABCD)

Huong
dan
gidi
a) Ta c6: MA + kMB = kMC
»
MA = k(MC
-
MB)
«
MA = kBC
. y.
Vay
tap
hop diem
M la
duong thang
di qua A va
song song
voi
canh
BC
ciia AABC.
b) Goi
J la
trung
la
diem AB,
I la
trung diem
JC ta c6

lA + IB + 2IC =
6
V
= MA + MB + 2MC= 4MI
Do do V cung phuong voi
BC M
thuoc
duong thang di qua
I va
song song
voi
BC.
Bai 1.55. Cho AABC. Tim tap hop diem
M
trong
cac
truong hop sau:
a)
2MA + 3MB 3MB-2MC b) 4MA + MB + MC
Huang
dan
gidi
a) Goi
K la
diem
thoa
man: 2KA + 3KB =
0
L
la

diem
thoa
man: 3LB
-
2LC =
0
Ta
c6:
2MA-MB-MC
2MA + 3MB
=
3MB + 2M(: o
MK
=
ML
=> Tap hop diem
M la
duong trung true cua doan
thSng
KL.
b) Voi
I la
trung diem ciia BC. Goi
J la
diem
thoa
man:
4JA
+
JB

+
JC
=
6
Ta
c6:
4MA + MB + MC
2MA-MB-MC
6MJ
2MA-2MI
6MJ
2IA;
1
<^MJ = -IA
3
Vay tap hop diem
M la
duong tron tarn
J
ban kinh
R
=
-
lA
.
3
Bai 1.56: Cho
iu
giac
ABCD

.
a) Xac dinh diem
O sao
cho
:
OB +
40C
= 20D
,
b)
Tim
tap hop diem
M
thoa
man
he
thuc
IMB
+ 4MC
-
2Mi3
=
3MA
Huang
dan
gidi
a) OB +
40C
= 20D <r> OB = -CI vol
I la

trung diem BD
3
b)
MB
+ 4MC
-
2MD
=
3MA| IMO
=
MA
Vay tap hop diem
M la
duong trung true ciia doan
OA.
Bai 1-57: Cho luc
giac
deu
ABCDEF
.
Tim tap hop
cac
diem
M sao
cho
:
MA
+
MB
+

MC
+
MD
+
ME
+
MF
nhan gia tri nho nha't.
Huang
dan
gidi
Ggi
P la
trong tarn ciia
AABC,
Q la
trong tam cua
ADEF
.Ml":
MA
+ MB + MC
MD
+ ME + MF
=
3
MP
+
3
MQ
=3(MP + MQ)>3QP

Dau
"
=
"
xay
ra
khi va chi khi
M
thuoc
doan PQ
Vay tap hop
cac
diem
M
can tim
la
moi diem
thuoc
doan PQ
Bai 1.58: Tren
hai tia Ox va Oy
ciia
goc xOy lay hai
diem
M, N sao cho
OM
+
ON
= a voi
a la

so' thuc cho truoc. tim tap hop trung diem
I
ciia doan
thang
MN
Huang
dan
gidi
Goi
hai
diem M„,
N„
Ian
lugt
thuoc
tia Ox va Oy sao cho
OMQ
= ON(, =

Gia
sir
OM
= k,
0
<
k < a khi do
ta c6
MI = ^^MoN(,'.
2
a

Do do tap hop diem
I la
doan
MQNQ
.
DANG
TOAN 7:
XAC
DINH TINH CHAT
CUA
HINH
KHI
BIETMOTDANG
THUC VECTO
1.
PHLTONG PHAPGIAI.
Phan
tich
dugc
djnh
tinh
xua't phat tir
cac
dSng
thirc
vecto
ciia gia thiet, luu
y
toi nhirng
he

thuc
da
bie't
ve
trung diem ciia doan thang, trong tam ciia
tam giac
va ket
qua " ma + nb =
0
<=>
m
=
n
=
0 voi a, b la hai
vecto
khong
Cling
phuong".
s , ^
2.
CAC VI DU.
Vi
du
1:
Goi M,
N
Ian lugt
la
trung diem

cua cac
canh
AD
va
DC ciia tir
giac
ABCD.
Cac
doan th^ng AN
va BM ck
nhau
tai P.
Bie't
?U
=
^BM;
2
-
AP = —
AN.
Chung minh rang tir
giac
ABCD
la
hinh binh hanh.
5
Lai
gidi
Taco:
AB = AM +

i^
= AM + 5MP
43
= SAP - 4AM = 2AN - 2AD
= 2(AD + DN) - 2Al3
= 2DN = DC
=i>
ABCD la hinh binh hanh.
Vi
du 2: Cho tam
giac
ABC c6 cac
canh
bang
a, b, c va trong tarn G
thoa
man:a^GA
+ b^GB + c^GC = O.Chung minh rang ABC la tarn
giac
deu.
Lai
gidi
G
la trong tarn tam
giac
ABC nen GA + GB + GC = 0
<=>
GA = -GB - GC.
Suyra
a^GA + b^GB

+c^GC
= 6.
a^ (-GB - GC) + b^ GB + cGC = 0.
Ofb
o(b2-a2)GB
+
(c2-a2)GC
= d.(*)
' Vi GB va GC la hai
veco
khong ciing phuong, do do (*) tuong duong voi:
fb2-a2=0
• c:>a = b = c hay tam
giac
ABC deu.
c -a =0
Vi
du 3: Cho tam
giac
ABC c6 trung tuyen AA' va B' , C la cac diem thay
doi
tren CA, AB
thoa
man XK' + BB' + CC' = 0. Chung minh BB', CC la
cac trung tuyen cua tam
giac
ABC .
Led
gidi
Gia six AB' = mAC, AC' = nAB

Suy ra BB' = AB' - AB = mAC - AB
va CC" = AC' - AC = nAB - AC
Mat
khac
A' la trung diem ciia BC nen AA' =
^(AB
+ Ac)
Dodo AA' + BB' + CC = d <::>-(AB +AC) + mAC-AB + nAB-AC = 0
f
1^
f 1^
n
— AB +
m
—
AC
= 0
I
2)
I 2)
Vi
AB, AC khong cung phuong suy ra m = n = - do do B', C Ian lugt la
trung
diem ciia CA, AB
Vay BB', CC la cac trung tuyen ciia tam
giac
ABC .
3. BAI TAP
LUY^N
TAP.

Bai 1.59: Cho tu
giac
ABCD c6 hai duong
cheo
cat nhau tai O
thoa
man
OA + OB + OC + OD = 0. Chung minh tu
giac
ABCD la hinh binh hanh.
Cty TNHH MTV DWH
Kliang
Viet
Huang
dan
gidi
Dat OA = xAC, OC = yAC, OB = zBD, OD = tBD
Suy ra 6A + OB + OC + OD = 0»(x + y)AC + (z + t)BD = 0
Do do
X
= -y; z = -t
=>
OA = OC, OB = OD nen tu
giac
ABCD la hinh binh hanh.
Bai 1.60: Cho ABC c6 BB', CC la cac trung tuyen. A' la diem tren BC
thoa
man
AA+BB'+OC=0.
Chung minh AA' cung la trung tuyen ciia tam

giac
ABC .
Huang
dan
gidi
Taco
AA' + BB' + CC' = 0
BA + BC CA + CB • ^r-, -
oAA
+ + =
0<=>BA
+CA =0
2 2
<=>
AA' cijng la trung tuyen ciia tam
giac
ABC .
Bai 1.61: Cho ABC c6 A', B', C la cac diem thay doi tren BC, CA, AB sao cho
AA',
BB', CC dong quy va
thoa
man AA' + BB' + CC = 0 Chung minh
AA',
BB', CC la cac trung tuyen ciia tam
giac
ABC .
Huong
dan
gidi
Gia su A'B = kA'C, B^ = mB^A, C A = nC'B

A^
= kA~C <^ AB - AA' = k(AC - AA') AA' =
Tuang tu ta c6
^,
BC-mBA
(m-l)AB + AC ^ - CA-nCB -nAB + (n-l)AC
AB-kAC
1-k
Aid.
1-m 1-m
AA'
+ BB' + CC' = 0
1
Suy ra
n
1
1-n
AB +
-k
1-k 1-m
-1
1-n
AB = 0
1-n
-k
1
-l=0=>k = m = n.
1-k 1-n 1-k 1-m
Mat
khac

theo
djnh li
Xeva
ta c6 kmn = -1 nen k = m = n = -1
Vay AA', BB', CC la cac trung tuyen ciia tam
giac
ABC .
Bai 1.62: Cho 4 diem A, B, C, D; I la trung diem AB va J
thuoc
CD
thoa
man
AD
+ BC = 2fi. Chung minh J la trung diem ciia CD. ^^^^
Huang
dan
gidi
AD
+ BC = 2ij<:^ID + IC = 2IJ
Goi K la trung diem DC suy ra 10 + 10 = 2^ do do K = J hay J la trung
diem ciia CD.
Bai 1.63: Cho ti>
giac
ABCD . Gia sit ton tai diem O sao cho OA = OB = OC = OD
va OA + OB + OC + OD = d . Chii-ng minh
rang
ABCD la hinh chu nhat.
Huang
dan
gidi

Goi M, N, P, Q la trung diem cua AB, BC, CD, DA A
Tit phuong
trinh
thu hai ta
duoc:
0 = OA + OB + OC + OD = 2OM + 2OT M'
oOM
+ OP-d
<=> M,P,0 thang hang va O la trung diem MP
B
N
Hinh
1.55
'AA«
0
= OA + OB + OC + OD = 2ON + 2C)Qc:>ON + OQ = 0
<=> N,Q,0 thang hang va O la trung diem NQ.
Ta
CO
AOAD can tai O nen NQ 1 AD, AOBC can tai O
nen NQ 1 BC suy ra AD
/
/BC .
Tuang tu AB//DC suy ra ABCD la hinh binh hanh
Ma N, Q la trung diem ciia BC, AD nen AB
/
/NQ => AB 1 BC
Suy ra ABCD la hinh chu nhat.
Bai 1.64: Cho tam
giac

ABC noi tiep duang tron tam O, goi G la trong tam tam
giac
ABC . A', B', C la cac diem thoa man:
OA = 30A', OB =
BOBOC
= 30C'.
Chung minh
rang
G la true tam tam
giac
A'B'C.
Huang
dan
gidi
G
la trong tam tam
giac
ABC nen 30G = OA + OB + OC
Do do OG = OA' + OB' + 6C'
Suy ra G la true tam tam
giac
A'B'C
Bai 1.65: Cho tam
giac
ABC noi tiep duang tron tam O, goi H la true tam tam
giac.
A', B', C la cac diem thoa man: OA' = 30A, OB' = 30B,
OC"'
= SOC.
Chung minh rang H la trong tam tam

giac
A'B'C
Huang
dan
gidi
H
la true tam tam
giac
ABC suy ra OH = OA + OB + OC
Do do 30H = OA' + OB' + OC hay H la trong tam tam
giac
A'B'C.
Bai 1.66: Cho tam
giac
ABC va diem M nam trong tam
giac.
Duang th^ng AM
cat BC tai D, BM cSt CA tai E va CM c3t AB tai F. Chung minh
rling
neu
AD
+ BE + CF = 0 thi M la trong tam tam
giac
ABC.
Huang
dan
gidi
Truoc tien ta chung minh bo de sau
Cho ba vec to a; b;c doi mot khong cung phuong va thoa man dieu kien:
ma + nb + pc = 0 (^,^,^0)

m'a +
n'b+p'c
= 0 ft,
Chung minh
rSng
•
= A =
•
That vay :
pi
thay m, m' * 0 thi suy ra ngay n, n', p, p' cung phai
khac
khong.
Tu
gia thie't ta c6 :
• n
, •
p •
a = —b + -i c
m
m
(3 I'l
a =
n
b
+
m
m
Vi
mot vec ta chi phan tich dugc mot

each
duy nhat qua hai vec to khong
nn'pp'm_n_p
cung phuong nen _ = _;_ = _=>- = - = -
Tro lai bai toan
— _ . . An • RF • CF •
Taco
AD + BE + CF = Oo
iliiMA
+ —MB + —^MC = 0
Mat
khac
Tuang tu
MA
MB
MC
I
k
1
MA
S^BM
SACM
S-S/
, CF
va
_BE
MB~S-S|,
"'^ MC S-S^
(vai S = S^Bc' Sa = SMBC ' - SMCA ' - SMAB )
^ , MA MB MC -

Dodotaco^.^. —= 0
Mat
khac
ta cung c6 S^, MA + MB + S, MC = 6
1
1 1
S-S
. . S-S, S-Sb
Ap dung bo de suy ra ' = —
trong
tam tam
giac
ABC .
^ e> Sg = Sb - S^ hay M trung
II
.Pi
DANG
TOAN 8:
CHUNG
MINH BAT
DANG
THUC
VA TIM
cue TBI LIEN QUAN DEN DO DAI
VECTO
•
PHUONG PHAP.
Su dung bat
d§ng
thuc ca ban:

Voi
moi
vecto
a, b ta luon c6:
+
+
a + b
a-b
+
, dau
bSng
xay ra khi a, b cijng huong
, dau bang xay ra khi a, b
nguoc
huong
47