toán nâng cao và các chuyên đề hình học 9

toán nâng cao và các chuyên đề hình học 9 tham khảo

P h i i n th* "hiit

Chudng I

61 MOT ~6 HE THUC LUQNG VE CANH VA D U ~ N G CAO TRONG

TAM GIAC V U ~ N G

- - Trong mot tam g i i c vuBng, binh

phuang mdi canh g6c vuang bing tich c i ~ a

canh huy&n v6i hinh chi& cua canh g6c

vuBng do tr&n canh huy&n

b = ab'; c = act RA -_ - - _ H _ _ - - - _ _ c

- Trong m&t tam g i i c vudng, binh -_ -

B

phuung d u h g cao irng v6i canh huy&n Hit~lz I

bhng tich hai hinh chi& cha hai canh g6c

vuBng tren canh huy&n

~ r o n g mot tam g i i c vuBng, tich hai canh g6c vuBng bing tich c i a

canh huy&n v6i duirng cao tuung irng

Trong m&t tam gi6c vuBng, nghjch dao binh phuung d u h g cao img

v6i canh huy&n bing tBng c i c nghjch d i o binh phuung hai canh g6c

_ G i i i

Gih sir tam giiic ABC vu6n&% , c6

KA AH I BC Theo he th* lu(mg v6 canh g6c vubng v6i hinh chi&u cha n6

tr&n canh huy&n, ta c 6 :

Vi & 2 MQt tam giiic vubng c6 canh huy&n

I i 6,15cm, d u h g cao h g v6i canh huy&n l i

3cm Tinh cdc canh g6c vubng cha tam gihc

Gitii

BC = 6.15 cm v i d u h g cao AH = 3 cm (h3) B -.- - 11 c

Theo he thirc l u w g v& d u h g cao v i hinh chi&u 6-15 - - * -

cha c i c canh g6c vubng tr&n canh huy&n, ta c6 : Hirrh 3

AH^ = BH HC

hay 32 = BH(6,15 - BH), suy ra B H ~ - 6,15BH + 9 = 0

o (BH - 3,75) (BH - 2,4) = 0

G i i su AB < AC, thg thi BH = 2,4 cm, khi d 6 HC = 3.75 cm

Cting theo h& thirc luqng trong tam g i L vubng ABC, ta lai c 6 :

2 Cho tam g i i c ABC vuang b A, dubng cao AH Bigt BH : HC = 9 : 16,

AH = 4 8 cm Tinh do d i i c i c canh g6c vubng cua tam giic

3 'Trong met tam g i i c vuang, ti s 6 giCia dubng cao v i trung tuygn kk ttir dinh g6d vubng bhng 40 : 41 Tinh do d i i c i c canh g6c vuang cua tam

g i i c vubng 66 bi6t canh huy@n bhng & Icm

4 Cho hinh vubng ABCD v i diim 1 nitn giiia A v i B Tia DI cat BC b E

D u h g thing kk qua D vuang g6c viri DE c8t BC b F

a) Tam g i i c DIF l i tam g i i c gi ? Vi sao ?

b) Chimg minh ring -;- +- khbng ddi khi I chuy4n dong tr&n

DI- DE'

doan AB

5 Cho tam g i i c ABC vubng b A c6 canh AB = 6 cm, BC = 10 cm CBc

d u h g phin g i i c trong v i ngoii cua g6c B cat AC Ian luut b D v i E

Tinh c i c doan thing BD v i BE

6 Cho tam g i i c ABC vubng b A, phin g i i c AD, dubng cao AH Big1

CD = 68 cm, BD = 5 1 cm Tinh BH, HC

7 Cho tam g i i c nhon ABC, hai d u h g cao BD v i CE c i t nhau tai H Goi B I , C I l i hai dikm tuang irng tren c i c doan HB, HC Bigt -~ ~-~ AB,C = AC,B = 90'

Tam g i i c ABICl l i tam g i i c gi ? Vi sao ?

8 Canh huykn cha mot tam g i i c vubng 1611 hun met canh g6c vubng c i ~ a tam g i i c 1 i 9 cm, chn tdng hai canh g6c vubng 1131 hon canh huyen lh

6 cm Tinh chu vi v i dien tich cua t a m g i i c vubng d6

9 Chci tam gi5c ABC c d ; 90' d u h g cao BH Dgf BC = a, CA =.b, AB = c,

AH = c', HC = b'

1

Chimg minh ring a = b + c = 2bc'

10 ChotamgiicABCc6 B = 6 0 ° , A C = 1 3 c m v i B C - B A = I

Tinh dQ dai cilc canh AB, BC

11 Cho tam giilc ABC c6 > 90°, d u h g cao j 3 ~ Dst BC = a, C

A C = c , A H = c ' , H C = b '

Chimg minh ring a2 = b2 + c2 + 2bcq

12 Cho tam giic ABC can ir B v i di6m D tren canh AC Bi&t BDC = qO,

AD = 3 dm, DC = 8 dm Tinh d& dii canh AB

$2 ~f SS LUQNG GIAC CUA GOC NHON

- d6i slna = - ; cosa = - k& ,

N6u hai g6c nhon a v i + c6 sina = sinp C canh huyen B

fhoac cosa = cosp, hocc ttga = tgp, ho#c Hl1ih4

cotga = cotgp) thi a = p

N6u hai g6c phu nhau thi sin g6c niy bang casin g6c kia v i tang g6c

niy bing catang g6c kia

N 6 u a + + = 9 0 ° t h i :

sina = cosp ; cosa = sin0 ;

tga = cotgp ; cotgp = tga

Vi du 3 Cho tam gi6c ABC vuang ti A, d u h g cao AH Bi6t AB = 7.5

Tam giic ABC vuOng 6 A, c6 AH I BC,

theo he th~Ic luqng trong tam giic vuang, ta c6 : Hi~rlr 5

45 Hai tam gidc vudng ABC v i A'B'C ddng dang viri n h a u ( i = - 900) c6 hai dubng cao h h ' tuung irng thuec canh huyen a v i a'

Chirng minh SHTK = ( I - cos' A - cos B - C O S - C ) S ~ ~ , ~

47 Cho tam giAc ABC vu8ng u A

a) K t dubng cao AA' ~ o i E v i F the0 thir tu 18 hinh chi& cha didm A'

CE AC"

tren AC v i AB Chirng minh - = -

BF , 4 ~ ~ ' b) ~ h o D I i met didm tr&n canh BC ; M v i N Ian Iuqt I i hinh chi&u c i a di6m D tren AB v i AC Chirng minh DB DC = MA MB + NA NC

48 Tren met qua doi c 6 mat c l i thdp B

cao loom Tir dinh B v i chin C c i a r - - - -

thip nhin diim A b chgn ddi duiri cic

g6c tuung h g bang 60' VA 3 (11 11 )

Tinh chi& cao h ciia quH ddi

49 6 dB cao 920 m, tir met m6y bay

truc thHng nguiri ta nhin hai diem A

v i B cha hai d%u met

cha met d&t c i c g6c ldn luqt IP H i ~ r h I J

a = 3 7 ' v ? i P = 3 1 °

Tinh chi& dhi AB c6a cdu

50 Cho tam g i i c AMB vudng b M Qua B k6 dubng thing d vu8ng g6c v6i

AB Goi H v i K ldn luqt l i hinh chi& c c l diem M tr&n.duting thing d

v i tr€n AB Cho bi6t &IAB = a ( a 5 459 vi AB = 2a

a) Tinh MA, MB, MH the0 a vh a ;

b) Tinh MH the0 a v i 2 a ;

C) Chirng minh : cos2a = 1'- sin a , cos2a = 2cos a -

5 1 DINH NGHfA VA S u XAC DINH D U ~ N G T R ~ N

T&p hap cbc didm clch didm 0 cd djnh met khoang bang R kh6ng d6i (R > 0) 11 d u h g trbn tam 0 c6 bbn kinh bang R

D u h g kinh 11 day cung l6n nhgt c6a dubng trbn

Qua ba di&m kh6ng thing hhng, bao gib ciing v8 duac met d u h g trbn

v i chi m6t m i th6i

D u h g trbn di qua ba dinh A, B, C c i a tam gibc ABC goi l i d u h g

trbn ngoai ti6p tam gilc ABC Khi 66 tam gibc ABC goi lh tam gibc n6i ti6p d u h g trbn

V i & 6 Cho tam gibc ABC vu6ng b B, AB = 8cm, BC = 6cm Goi D 11

didm d6i ximg c6a didm B qua AC

a) Chirng minh r&ng b6n didm A, B, C, D chng thuoc met dubng trbn b) Tinh bbn kinh cba d u h g trbn n6i trong

Gicii

a) Theo d& bhi, didm D d6i ximg v6i

didm B qua AC n6n DA = BA, DC = BC

AADC = - AABC (c.c.c), -

suy ra ADC = ABC = 90"

Goi 0 lh trung didm cba AC Trong cbc

tam g i k vu6ng ABC v i ADC c6 BO v i DO

IP cbc d u h g trung tuy6n thuec canh huy&n

AC n6n BO = OA = OC, DO = OA = OC

Suy r a : O A = O B = O C = O D

V$y b6n didm A, B, C, D c h g thubc

dutfng trbn tam 0 d u h g kinh AC Hinh 12

b) Tam gihc ABC vu6ng b B, the0 djnh l i Py-ta-go, ta c6 :

2 2

AC = AB + AC' = g2 +62 = 100, suy ra AC = LO cm

' a) Cich dung :

I

- D;mg dubng trung trqc d c6a

doan AB, d c i t tia Ay b 0:

- Dung d u h g trbn t8m 0 , bin

kinh OA thi dubng trbn ( 0 ) chinh Ih

d u h g trbn phai d ~ p g

Chrhry nrinh :

truc d c i ~ a doan AB n&n OA = OB, do

V$y bsn kinh d u h g tr6n (0) 1B : R ,= -cm

2

51 Cho tam g i i c d$u ABC, hai d~rimg cao BD, CE

a) Chimg minh b6n di&m B, C, D, E chng thuec met d u h g trbn ; b) Goi G 18 giao diim cha BD vh CE Chirng minh b6n didm A, E, D, G chng thu&c m6t d u h g trbn Tinh bin kinh cba d t h g trbn niy, bi&'t tam gi6c d&u ABC c6 canh bing 8cm

TOAN NANG CAO HlNH 9 - 2

; I

52 Cho d u h g trbn ( 0 ) day AB V day BC vu6ng g6c viri ABI

a) Cheng minh AC I i dudng kinh cha d u h g trbn ( 0 )

b) Tinh b i n kinh d u h g trbn (O), bi6t AB = 12cm BC,= 5cm:

53 Cho hinh vubng ABCD Tren c i c canh AB BC, CD, DA My theo thir tu

AM BN CP Q D

c i c didm M, N, P, Q sao cho - = - = - = -

MB NC PD Q A ' a) Chirng minh tit g i i c MNPQ l i hinh vu6ng ;

b) Goi 0 l i giao didm hai dudng chCo AC v i BD C h h g minh ban d i d ~ n

M, N, P, Q cfing n i m tr&n mbt dubng trbn tam I i diim 0

54 Cho d u h g trbn tam 0 bin kinh 4cm, c6 tam b goc toa do H i y x i c d!nh v/ tri c6a c i c diim A, B, C dbi v6i duirng trbn, b16t toa d o clic di6m :

A ( - 2 ; ) ; B 6 ) ; C ( 2 6 ; - 2 4 3 )

55 Goi I v i K theo thir t$ I i c i c diem nim tr&n canh AB, AD cua hinh vubng ABCD sao'cho A1 = AK Dubng thing k ~ ? qua A vu6ng g6c vCli

DI b P, c L BC b Q

C h h g minh nBm didm C, D, K, P, Q cilng thuec mot dubng !rbn

56 Cho hinh vu6ng ABCD, 0 I i giao didm hai dubng chCo AC v i BD Goi

M, N 1611 IU@ I i trung didm c i a OB, CD -

a ) C h h g minh AMN = 90°, tir d6 suy ra bb'n didm A, M, N, D c i ~ n g thuqic mqit d u h g tr6n ;

b) So s i n h AN viri MD

57 ~ h o tam g i i c ABC can 6 A c 6 AB = IScm, dubng cao AH = 9cm

Tinh b i n kinh d u h g trbn ngoai tisp tam g i i c ABC

58 Cho hinh vusng ABCD canh bdng a Goi M v i N I i hai d i i m tujr 9 tr&n

c i c canh AB v i AD sao cho chu vi tam g i i c AMN bing 2a Goi H 18

hinh chi& cua didm C tren MN

Chirng minh rang didm H lubn lu6n thuqic mqit d u h g trbn c 6 djnh khi

hai diim M, N chuybn dong tr&n c i c canh AB v i AD

59 Cho dubng trbn ( 0 ; R) v i m6t di6m A nam trong d u h g trbn d 6 (A khlic 0) Tim tap hqp trung didm M c i a doan thing AB khi didm B chuyin dong tren dubng trbn (0)

Tam cua dubng trbn Is ihm d6i xirng c i a d u h g trbn 66

Bgt ki dudng kinh n i o cting ,i truc dPji xitng c i a d u h g trbn I

D u h g kinh vudng g6c v&i met day cung thi di qua tnrng dikrn c6a day gy I

D u h g kinh di qua trung didm cua met day (kh6ng 18 dubng kinh) thi vuang g6c vdi day &y

Trong met d u h g trbn :

- Hai day bang nhau thi c l c h d&u tam'

- Hai day cdch dku tam thi bang nhau

-Day Ian han thi g l n tam hun

- Day g l n tam hun thi ldn hun

V i [/<I 8 Cho tam giic ABC nei ti+ dubng trbn ( 0 ; R) ; Tinh dB dhi cdc canh AB, AC, bik't R = 3cm vh khoang cich tir 0 den AB vh AC Iln luut 18

Bdi t@p

60 Cho d u h g trbn ( 0 ; R ) hai dtly cung AB v i CD c&t nhau tai diim M

n i m &n ngoii d u h g trbn

, a) Chimg minh ring n€u AB = CD thi MA = MC ;

b) T r u h g hqp AB > CD HHy so sinh khoing cich til M d&n trung diim cha c i c dtly AB, CD

61 Cho d u h g trbn ( 0 ) v i didm I nim b&n trong dubng trbn Chirng minh rhng trong c i c dtly di qua I thi dily vuBng g6c vdi 0 1 c6 dB d i i nho nhst

62 Cho nira d u h g trbn ( 0 ) dubng kAh AD Tr&n nira dubng trbn l&y hai di&m B v i C Bi8t AB = BC = 2&cm, CD = 6cm Tinh b i n kinh d u h g trbn

63 Cho dubng trbn (0 ; R), d u h g kinh AB, dtly cung AC

a) Cho bi&t khoang cich tir 0 d8n AC, BC l8n IU@ l i 6cm v i 8cm

Tinh d& d i i c i c dily AC, BC v i bin kinh d u h g trbn ;

b) Tr&n tia d6i c i a tia CA Igy diim D sao cho CD = CA Tim tap hqp , trong tilm G ctia tam giic ABC khi C chuydn dong tr&n d u h g trbn ( 0 )

64 Cho dubng trbn ( 0 ) d u h g kinh AD, dily cung AB Qua B k6 day BC '

vuBng g6c vdi AD

I Tinh b i n kinh cba d u h g trbn bigt AB = IOcm, BC = 12cm 1

65 Cho d u h g trbn ( 0 ) bin kinh 5cm Hai dtly AB v i CD song song v6i nhau

v i c6 do d i i l8n luqt bing 8cm v i 9.6cm Tinh khoing cich giila hai dtly

66 Cho d u h g trbn ( 0 ; R), dubng kinh AB, day cuhg DE Tia DE c6t AB

b C Bi&t DOE = 90' v i OC = 3R

a) Tinh d o d i i CD v i CE the0 R ;

b) Chimg minh CD.CE = CA.CB

67 Cho d u h g trbn (0.; R ) v i mot diim M nim b8n trong dubng trbn

a) Hgy n&u cich dqng dtly AB nhan didm M Iim trung dl& ;

b) Tinh d& d i i day AB n6i trong ctlu a bi€t R = 7,5cm, OM = 2, lcm I

68 Cho nira d u h g trbn ( 0 ; R) Hai dily cung AB vA CD song song viri nhau c6 do d i i 18n luat l i 32cm vh 24cm v i khohng giaa hai dtly I i 4cm

Goi d l i khoHng cich tir tam 0 cha d u h g trbn ( 0 ; R) d&'n d u h g thing a thi vj tri tuang d6i cha d u h g thing v i d u h g trbn duqc bidu thj the0 bHng sau :

3 D u h g thing v i d u h g trbn kh6ng giao

Vi tri tuung d6i cha '

d u h g th&ng v i duirng trbn

I D u h g thing v i d u h g trbn tit nhau

2 Dubng thing v i d u h g t r b n ,ti&p x6c nhau

Vi du 9 Cho doan thing AB v i trung di&m 0 cha AB Tr&n tang mot nira m+t phing bir AB vZ tia Ax, By vu6ng g6c vdi AB Tren cBc tia Ax, By Igy theo thir tu hai dikm C v i D sao cho COD = 90' K6 OH I CD

a) Chimg m h h h g H thu6c d u h g trbn t h 0 ;

b) XBc djnh vj tri tucmg d6i cha d u h g thing CD v&i d u h g trbn ( 0 )

Gidi

a) Tia DO cht tia d6i cba tia Ax 13 E

Tam giBc vu6ng AOE v i tam g i i c vu6ng

BOD c6 :

S6didm chung

O D = O E (chirng minh tren) Hirth IS

Dl= E (chimg minh tren)

Do 6 6 AOHD = AOAE (canh huy&n - g6c nhon), suy ra OH = OA

Vi OA = OB m i OA = OH nen OA = OB = OH Vgy di&m H thu6c

d u h g trbn tam 0 d u h g kinh AB

b) Do H thuoc dubng trbn tam 0 bin kinh OA, mh C D I OH tai H n&n khoing cdch tir didm 0 d&n CD bing bin kinh d u h g trbn ( 0 ) nghya l i d =

R Vay d u h g thing CD ti6p xlic vdi d u h g trbn ( 0 ) tai didm H '

69 Cho dikm M cdch d u h g thing xy l I 6cm VB dubng trbn (M ; IOcm) a) C h h g minh ring d u h g trbn (M) c6 hai giao didm v6i d u h g thang xy b) Goi hai giao didm n6i tr&n II P vh Q Tinh do d i i PQ

70 Cho hinh thang ABCD.c6 = = 90°, AB = BC = lcm, AD = 2cm Chimg minh ring d u h g thing AC ti6p xlic vdi dubng trbn (D ; f i cm)

71 Cho hinh vuBng ABCD, tr€n d u h g chCo BD 18y didm I sao cho BI = BA Dubng thing k6 qua I vu8ng g6c vdi BD c i t AD 6 E

a) So s i n h c i c doan thing AE, EI, ID ;

b) Xbc djnh vj tri tucmg d6i c6a d u h g thing BD vdi d u h g trbn (E ; EA)

Ox v i c 6 tam n i m tr€n canh Oy

73 Dqng d u h g trbn b i n kinh 2cm, ti6p xuc v6i d u h g thing xy cho trubc

v i di qua diem A c i c h xy m8t khoHng 3cm

74 Cho d u h g trbn (0 ; 15crn), AB lh day cung c6a d u h g trbn Tim tap hqp trung dikm Vc6a AB khi AB thay d8i trong dubng trong (0), biQ

AB = 24cm

, 54 TIEP TUYEN CIIA D U ~ N G TRON

T ~ N H C H ~ T HA1 TIEP TUYEN CAT NHAU

Ti6p tuy€'n c6a d u h g trbn 1I d u h g thing chi c6 met dikm chung vdi

N6u hai tiEp tuy6n cc6 m&t duimg trbn cdt nhau tai m4t dikm thi :

- Didm 6 6 c i c h d&u hai ti6p didm

- Tia k t tir diem d6 di qua t&m 118 tia phan giic cha g6c tao biri hai ti&p tuy6n

- Tia k t ttr tam di qua di6m 66 lh tia phrln giic cba g6c tao biri hai b i n kinh di qua ti&p dikm

! V i tlg 10 Cho dubng trbn (0 ; 5cm) vh didm M nhm b6n ngohi d u h g

trbn Qua M k& hai ti&p tuygn MA vh MB vdi d u h g trbn (A vh B lh tigp

1 dikm)

Ti3 di&m C tr6n cung i1h6 k& tigp tuygn v6i d u h g trbn d t MA, MB IPn luqt 6 P vh Q Cho bi&'t AM I BM

a) Tir gi6c'MAOB 11 hinh gi ? Vi sao 7

b) Tinh chu vi tam giic MPQ - ;

C) Tinh g6c POQ

Gidr

a) Theo giH thigt :

MA, MB 18 tigp tuy&n cha d~rbng trbn ( 0 )

c3 , A vh J3 n6n OA 1 AM vh OB 1 BM, Go d6

MA I MB n t n G = 90°

Tir g i i c AMBO c 6 = = = 90' ,

n&n 18 hinh chii nhgt Hinh chii nhgt nhy lai

b) Theo tinh chgt hai ti6p tuygn cc6 mot Hinh 16

I 1 0

= -AOB = - .90 = 4

Vi d41 I I Cho tam g i i c ABC ctin Zr A VE d u h g trbn t i m D d u h g kinh

BC c k AC v i AB Iin l u g Zr E v i F Goi H 18 giao didm cba BE v i CF

C h h g minh ring :

a) B6n digm A, E H, F tang thuec met d i r h g trbn

b) DE 18 ti&p tuy&n c i a dubng trbn n6i trong ciu a

Goi 0 l i trung di8m cha AH, ta c6 OE,

O F I&n I ~ o t i i trung tuy&n thuec canh huy&n

AH cGa hai tam giic vu6ng AEH v i AFH n&n

OA = OH = O E = OF Do 66 b6n di&m A, E,

H, F thucc dubng trbn (0)

v i CF cba tam g i i c ABC, AD l i dubng trung Hinh 17

tuy&n thuQc canh d i y BC n&n AD I BC, d o

suy rz + E2 = 90' hay O ~ D = 90' hay OE I DE

DE vu6ng g6c v6i bin kinh OE tai E n&n DE is ti&p tuygn cha d u h g tmn (0)

75 Cho dubng trbn di&m tr&n d u h g trbr sao cho BAC ( 0 ; 5cm) dubng kinh AB, ti&p tuy&n Bx Goi C l i met - = 30 0, tla AC cfd Bx . b E a) C h h g minh BC' = AC CE ; '

b) Tinh d o d i i doan BE

76 Cho nha d u h g trbn (0) d u h g kinh AB Qua C thuac nka d u h g trbn, k6 tiCp tuy6n xy cua nha dubng trbn Goi M v l N Ian Iuut I i hinh chi& cba dikm A vb diim B tren xy Goi H lh than dubng vu611g g6c k6 tir C

a) C lh trung dikm cha MN ;

C h h g minh rhng :

1

MA + MB) < PQ <-(MA + MB)

79 Cho nira d u h g trbn ( 0 ) d u h g kinh AB, hai ti6p tuy6n Ax, By Trtn Ax,

By igy the0 - thit tu hai dikm C v l D Bi& AC + BD = CD Chimg minh : a) G6c COD = 90'

b) D u h g theng AB lb ti6p tuy&n c6a d u h g trbn ngoai ti6p tam giic COD, cbn d u h g theng CD 1b ti6p tuy&n c6a dubng trbn (0)

80 Cho g6c xAy k h i c g6c bet Chimg minh ring c6 thk tim duuc v6 s b chc dikm B vb C tr&n hai canh Ax v l Ay sao cho chu vi tam g l l c ABC lu6n

luBn bhng 2 1 (1 l l mijt dij dhi cho trudc)

81 Cho d u h g trbn (0 ; 6cm) v l day AB bing IOcm Goi M 18 mot didm tren

d u h g thang AB vh M nam btn ngobi d u h g trbn (0) Tim khoing cich t& M d&n trung didm c6a AB khi g6c xen giiia hai ti&p tuy&n k6 tir M

bhng 60' ; 90'

82 Cho nira dubng trbn ( 0 ) d u h g kinh AB, ti6p tuy6n Ax Qua dikm C trtn nira d u h g trbn ( 0 ) k6 ti6p tuy6n vdi nira d u h g trbn cat Ax b M K6

C H I AB cdt BM b I C h h g minh I Ib trung dikm cba CH

83 Cho tam g i i c ABC c6 BC = IOcm, CA = 12cm v l AB = 14cm Tinh khoing c i c h giira tam d u h g trbn niji ti+ vh t r y g tam cha tam giic

84 Cho tam g i i c ABC c6 BC < AC, trung tuy&n CD D u h g trbn n6i tiep c8c tam giic ACD Vh BCD ti6p xxc vdi CD lln luqt b E vb F

C h h g minh : 2EF = AC - BC

85 Cho tam g i L ABC vu6ng ir A D u h g trbn ( 0 ) n&i ti&p tam gi6c ti6p x6c vdi canh AB, AC Iln luqt b D v l E

a) Tit g i L ADOE l l hinh gi7 Vi sao?

b) Tinh b i n kinh d u h g trbn (O), biB AB = 5cm ; AC = 12cm

86 Cho tam gi5c ABC, bigt BC = a, CA = b, AB = c Goi r l i bin kinh

, dtrirng trbn noi tigp, S 18 dien tich c i ~ a tam glac

Chirng minh ring : A B + AC = 2(r + R)

89 Cho tam gidc ABC vu6ng b A c6 BC = a, CA = b, AB = c Goi r lh bin kinh d u h g trbn n6i tiBp tam giic

Tinh do d i i c i c doan AE, AF, BE, CF

91 Cho tam g i i c d&u ABC Goi M, N la hai di&m lSin luqt tr&n hai canh AB,

AC ; D 18 trung di6m c6a BC

Bict chu vi tam g i k AMN blng nira chu vi tam gidc ABC Tinh g6c MDN

92 M6t tam g i i c vu6ng noi tiBp mot d u h g trbn duitng kinh 37dm v i ngoai tigp met d u h g trbn b4n kinh 5dm Tinh c i c canh g6c vu6ng cua tam gihc do

Goi 0, 0' 18 tam cha hai duirng trbn Dubng t h h g 00' goi 1 i d u h g noi tam, doan thing 00' goi Ih doan n6i tam D u h g n6i tam l i truc ddi ximg c6a hinh g8m hai d u h g trbn ( 0 ) v i ('0') ;

NBu hai d u h g trbn ti6p xlIc nhau thi tiBp di&m nim tren d u h g noi trlm

NBu hai d u h g trbn cat nhau thi d u h g nbi tam vu6ng g k vdi drly chung vh di qua trung dikm c6a drly chung

Hai d u h g trbn ( 0 ; R ) v i ( 0 ' ; r) c6 R 2 r Vj tri tuang doi giira hai

d u h g trbn img vdi he thttc giira R, r v i 00' duuc cho the0 bHng sau :

I , ' - Ti&p x6c ngohi I 1 d i h / 0 0 ' = R + r I

Vi tri tuong dbi cOa hai d~rbng

trbn ( 0 ; R) vh (0' ; r)

Hai d u h g trbn c i t nhau

Hai d u h g trbn ti6p xx6c nhau

Tr&n hinh 18, c i c d u h g thing d l , d2 Ih ti+ tuy6n chung ngoii ciia

hai dubng trbn ( 0 ) v i (0'), c i c d u h g thing m,, m2 18 ti6p tuy6n chung

trong c6a hai dubng trbn ( 0 ) v i (0')

S6 di6m chung 2di'm chung

R - r < 0 0 ' < R + r

V i dl( 12 Cho d u h g trbn ( 0 ) vh (0') tigp xuc ngohi tai A Ti6p tuy6n

chung ngohi cba hai d u h g trbn c6 ti6p di6m vdi & h g trbn ( 0 ) 6 M vdi d u h g

trbn (0') 6 N, ti&p tuy6n chung trong c6a hai d u h g trbn tai A c i t MN 6 I

a) Chirng minh tam giic MAN vh 010' 18 c i c tam g i i c vuBng ;

b) XBc djnh vj tri t u m g d6i c6a d u h g thing MN vdi d u h g trbn dubng

Suy ra IM = IA = IN, do 66 tam

giic MAN lh tam giBc vu6ng 2, A

?he0 tinh chgt hai ti6p ttuy&n c6a

mQt d u h g trbn ckt nhau, ta lai c6 : I 0 Hinh 19

vh 10' 1&n lugt lh tia phan g i k c6a hai

A

g6c k& bb MIA vh NIA , do 66 I 0 1 10' Vey tam giic 010' vuBng 2, I

b) Goi I' 18 trung diim c l a 0 0 ' , ta c6 1'1 = I 0 = 10' nen 1'1 18 bin kinh

d u h g trbn d u h g kinh 0 0 '

_ OM I MN vh O'N I MN ntn OM I/ O'N suy ra tir gihc OMNO' 18 hinh thang 1'1 lh d u h g k n g binh c i a hinh thang OMNO nen I' 1 //OM, suy ra 1'1 I MN

Dubng thing MN vuBng g6c v6i b6n kinh 1'1 tai I nen d u h g t h h g MN

lh t&p tuygn cc6 d u h g trbn (I)

V i (14( 13 Hai d u h g trbn (0, ; 6.5cm) vh (02 ; 7,5cm) giao nhau tai A

vh B Tinh dQ Cai doan n6i tam 0,02, bigt AB = 12cm

93 Cho nira dubng trbn ( 0 ) d u h g kinh AB VE d u h g trbn tam 0' d u h g kinh

OA Qua A vi5 d&y cung AC &a d u h g trbn ( 0 ) c i t d u h g trbn (0') b M

C h h g minh :

a) D u h g trbn (0') vh d u h g trbn ( 0 ) ti6p xxdc v6i nhau ;

b) O'M song song v6i OC ;

c) M 18 trung dikm c6a AC vh OM song song vdi BC

94 Cho tam giic ABC vudng a A VO d u h g trbn (01) di qua A vh tiEp x6c vdi BC tai B, vE d~rimg trbn (02) di qua A v i ti+ xxc vdi BC tai C Goi

M 1 i trung didm cSa BC Chdng midh :

a) D u h g trbn (0,) v i (02) ti&$ xx6c vbi nhau ;

b) AM I i ti+ tuyEn chung c6a hai d u h g trbn (0,) v i to2)

95 Cho hai d u h g trbn (0 - ; R ) v i (0' ; R') c i i nhau tai A v i B BiEt OAO'=~O", R = 6cm v i R' = 4,5cm

a) Tinh 00, AB ;

b) Goi P I i trung didm c6a 00, qua A kc5 c i t tuy&n vudng g6c vbi AP

c i t dubng trbn ( 0 ) if C, c&t d u h g trbn (0') 13 D So sdnh AC, AD v i AB

96 Cho hai dubng trbn (01 ; 17cm) v i ( 0 2 ; 1Ocm) AB I i mot tigp tuygn chung ngoii c6a hai d~rbng trbn c6 ti6p didm viri d~rbng trbn ( 0 , ) u A, vdi d u h g trbn ( 0 2 ) if B Dubng thing AB c i t dubng ndi tam 0101 b C Tinh d o &hi c i c doan C O I , C 0 2 bi6t 0102 = 21cm

97 Cho td g i i c ABCD Bi&i ring dubng trbn nOi tiCp hai tam giic ABC vh ADC ti6p xx6c nhau Chdng minh rgng d~rbng trbn noi tiEp hai tam giic ABD vh CBD c b g tiEp xx6c nhau

98 Cho hai d u h g trbn (0,) v i (02) ti6p xbc ngohi tgi A ~ d t dubng thing

ti+ xuc vbi d u h g trbn ( 0 , ) ir B, ti+ xxlic v6i d u h g trbn (02) if C BiEt

AB = 6cm, AC = 8cm

a) Tinh dQ dhi doan BC ;

b) Tinh b i n kinh c6a c i c d u h g ( 0 , ) vh (0.9 -

99 Cho hai dubng trbn ( 0 ) va (0') tiEp xhc ngohi u A Dubng nbi t2m 00'

c i t dubng trbn ( 0 ) b B, c i t d u h g trbn (0') b C DE 11 mot ti€$ tuy&n

chung ngohi cha hai dubng trbn (D E(O), EE(0'))

Goi M l i gino didm cua hai dubng thing BD v i CE Chirng mich : a) G6c EMD = 90' ;

b) MA l h tiEp tuy&n chung cda hai dubng trbn ( 0 ) v i ( 0 ' ) ;

c ) Tam g i k AMN' Ih tam gikc gi ? Vi sao ?

.' 101 Cho hai dubng trbn ( 0 1 ) v l (02) ti6p xuc ngohi U K AD 1l mot ti6'p

tuy6n chung ngoii' c6a hai dubng trbn ( A € ( O , ) , D E ( 0 2 ) ) VE dubng kinh AB c i ~ a dubng trbn (OI) Chfing m i n h ~ ~ ~ = BK.BD

102 Cho hai dulrng trbn (O1; 5cm) v i ( 0 2 ; 2cm) nim ngohi nhau Mot ti6p tuy6n chung ngoli AB c6a hai dubng trbn ( A € ( O I ) B E ( 0 2 ) ) vh met ti+ ttuy6n chung trong CD c i a hai dubng trbn ( c E ( o ~ ) , D E ( 0 2 ) ) Tinh do dbi doan n6i t8m O1O?, bi6t AB = 1.5CD

103 Cho d u h g trbn (0) vh d~rirng thing xy khong giao nhau v i hai didm

A, B trtn d~rbng thing xy Qua A vE hai ti6p truy6n AC vh AD vbi dnbng trbn (0) trong do C, D 12 c i c ti6p didm, qua B vZ hai ti@ tuy6n BE vh

BF vbi dubng trbn (0) tlong d6 E, F l i c i c tit5p didm VE d~rbng trbn ( 0 , ) ti6p xuc vbi hai canh AC, AD ; vE dubng trbn ( 0 2 ) ti6p xuc vdi hai canh BE, BF

C h h g minh ring n6u duirng trbn ( 0 1 ) v l (02) bhng nhau thi 0,02 song song v6i xy

104 Cho tam gikc cAn ABC c6 c q h d i y BC = 16cm, canh b8n AB = IOctn Tinh b i n kinh dubng trbn nbi ti66 ngoai ti6p tam gikc v l kholng cich giita c i c tam c i a hai d u h g trbn d6

105 Cho hai d u h g trbn (0) vh (0') giao nhau b A v l B ( 0 v l 0' thuoc hai nha mat phang bir AB) Met c i t tuy6n k e qua A tit dubng trbn (0) CI C

cat dubng trbn (0') u D K e OM I CD vh O'N I CD

I

a) Chdng minh MN = -CD ;

2 b) Goi I l i trung didm c i a MN C h h ~ g minh ring dubng thing k6 qua I vubng g6c vbi CD lubn ludn di qua mot diem c o d/nh khi c i t tuy6n CAD thay ddi ;

C) Qua A k6 ckt tuy6n song song vdi d u h g nbi tam 00' cat dubng trbn ( 0 ) b P, cat d u h g trbn ( 0 ' ) b Q So sinh do d j i ckc doan CD v i PQ

106 (Jho n h dubng trbn ( 0 ) d u h g kinh AB = 2R v i di6m M t&n n h d u h g trbn

66 VE d u h g trbn t h M ti6p xxilc vbi d ~ r h g kinh AB tai H Qua A vh B vE hai ti6p tuytn AC vh BD v6i d u h g t$n (M) trong 66 C, D lh ckc ti6'p didm a) Chdng minh ba didm C, M, D cbng nim tr8n ti6p tuye'n c i a dulrng trbn ( 0 ) tai diim M ;

C ) G i i s~ CD c i t AB b K Chirrlg minh OB' = O H OK

107 Cho 3 didm A, B, C theo thir t ~ r d6 trCn mot dubng thing v i AB = 4BC Trtn cung mot nira mat phang bir AC vE i ~ u a dubng trbn tam 0 dubng kinh AB v i nua dubng trbn t9m 0' doimg kinh BC Ti+ tuyCn chung cua hai nira duirng trbn co ti@ diCm viri duirng trbn (0) a F vdi nira dubng trbn ( 0 ' ) b G, c i t c i c 1i6p'tuyQ ve tit A v i C cua hai ~ i u a dubng trbn d6 b D v i E Ti6p ttuy&n chung cua hai n+ dtrirng trbn 6 B c i t DE

a ) C h h g minh c i c tam g i i c 0 1 0 , OID vh O'IE I2 c i c tam giic vuong ;

I b) Dat O'C = a ( a I.8 mot'do d i i cho truirc) Tinh 91, EG v$ AD theo a ;

I

c) Tinh - SADEC the0 a

1 Ms Cho, tam Ciic ABC c6 BC = a CA = b AB = b Gqi S I i di(n t k h cba

I

tam giic

I

C h h g minh rang ngu C(a + b + c)(b + c - a) = 4 s thi tam giic ABC vudng 6 A

109 Cho dubng trbn (0 ; R) v i duirng thing xy khdng giao nhau Tit mot didm M tujr ); tree xy kk hai ti6p tuy6n M P vA MQ vdi dubng trbn (0) trong d 6 P, Q ii ciic ti6p didm Qua 0 ke OH I xy, d8y cung PQ c i t OH

u I, cat OM b K Chirng minh :

1 b) PQ ludn luBn di qua mat di&m c6,djnh khi diem M thay d6i tr€n xy

v6i d u h g kinh AB c i ~ a d~rbng trbn ~ i i y tai M C h h g minh,AM BM = 2Rr

111 Cho duirng trbn (0 ; R) ti6p xxilc trong vdi d u h g trbn (0' ; R') tai didm

A trong d6 R' > R Duirng nbi t2m 00' cit hai duirng trbn niy idn l u g b

B v A B' C h h g minh rang ti6p tuy6n chung ngoii cha c i c dubng trbn

d u h g kinh 00' v i BB' di qua didm c b djnh A

112 Cho d u h g trbn ( 0 ) v i met didm P nkni btn tiong duirng trbn ( P + 0) Gpi Q l i met di&m tujl ); tr8n dirbng trbn ( 0 ) Chirng minh rang khi di6m Q chuydn dong tren dubng trbn (0) thi giao didm M ciic dubng thing k t qua 0 vudng g6c vdi PQ vh ti6p truyCn kC tit Q c i a duirng trbn ( 0 ) chay trtn met dubng thing c 6 djnh

113 Cho d ~ r b i g thing AB, hai diim C v i D thuoc hai ~ i u a mat p h i ~ ~ g khic nhau b b AB HHy tim trtn AB met di&m M sa6 cho AMC = 2BMD

114 Dvng hai d u h g trbn ti6p xilc ngoii c6 tam I.+ hai didm c 5 djnh cho trudc sao cho met trong hai ti6p tuy&l ch.ung 'ngoii cua liai dubng trbn

d 6 di qua ;not diem c b dinh cho trudc

C h z i ~ n g 111

GOC vdi D U ~ N G TRON

G6c c 6 dinh trirng viri t i m dudng trbn d u a c goi la g6c tam

S 6 d o c i ~ a cung nho bang s o d o g6c u t2m chan c u n g d6

S 6 d o c ~ i a riua dubng tr6n bang 1x0"

Trong m6t dubng iron hay hai duirng trbn b i n g r ~ h a u :

- Hai cung duqc goi I i bang nhau neu chung c 6 s 6 d o bang nhau

- Trong hai cung, cung n i o c 6 s o d o luri han d o a c goi la cung Ian han -

N6u C la mot dikm tren cung A B t h i :

O A = 0 1 ( b i n kinh d~rirng tron (0))

v i O'A = 0 ' 1 ( b i n kin11 duimg t r b ~ i

( 0 ' ) ) m i O A = O'A (theo gih thiet) n&n

O A = O'A = 0 1 = 0'1, d o d 6 tir g i i c OAO'I

la hinh thoi

Mat k h i c O A O ' = 90' (theo giii thie't) ~ I ~ I I / J 21

vi the" td g i i c OAO'I Is hinh vu6ng

-

b) Td g i i c OAO'I la hinh vuBng n&n AOI = A O ' I = 90"

Trong d~rirng trbn ( 0 ) g6c b tilm A 0 1 = 90", d o d 6 :

- S o d o c u n g n h o A1 = s 6 d o g6c A 0 1 = 90"

- S 6 do c u n g 1611 A1 = 360' - 90" = 270'

Tuang tu, trong dubng trbn (O'), ta c6 :

- Cung l6n A1 c6a d u h g trbn ( 0 ) = cung Ian A1 c6a d u h g trbn (0')

115 Cho dubng trbn ( 0 dubng kinh BOC vh BOD So sanh so" do c i a hai-cung nh6 AC v l AD ; R ) vh d u h g trbn (0' ; R'), c i t nhau tai A v l B - - VE

trong hai d u h g trbn d6, bigt R R R'

116 Cho tam giic d&u ABC VC phia ngoli cua tam g i i c v& nira d u h g trbn tam 0 , dubng kinh BC Tr&n nira dubng trbn d6 lily hai di6ni M v l - - N sao

-cho cung BM = MN = NC Goi giao di6m c6a AM, AN vdi canh BC IBn lugt 18 D v i E

Chting minh BD = DE = EC

117 Cho d u h g trbn ( 0 ; R) Tren ti&p tuy&n kkt tir A v8i dubng trbn l&y didm B, tia OB c i t d u h g trbn ir C Bi&t AB = R& Tinh s15 do c i c cung

AC c l a d u h g trbn ( 0 ) , (Ilm trbn d&n do)

118 Cho hai dubng trbn ( 0 ; R) v l d u h g trbn (0' ; R) c i t nhau tai A v i B trong 66 tam c l a d u h g trbn nly nim tren d u h g trbn kia

a) Tir g i i c AOBO' I l hinh gi ? Vi sao ?

n b) Tinh do d l i c i c cung AB c6a m6i dubng trbn ;

c) SS d o hai cung nhb bang nhau

120 C h o tam g i i c OBC can b 0, dvbng cao OH VE d u h g trbn (0 ; OB), tia H O c i t dubng trbn 0 h A Bigt b = 50''

Tinh s6 d o c i c c u n b nh6 BC, CA, AB

52 LIEN HE G I ~ A CUNG V A DAY

! Vdi hai cung 11116 trong mat d ~ r i m g trbn hay hai duirng trbn b i n g nhau : I

1 - Hai cung bang nhau cang hai day bang nhau I / - Hai day b i n g ntiau dang hai cung bang nhau I

I Vdi hai cung nh6 trong m6t duimg trbn hay hai @ u h g trbn bang nhau :

- Cung ldn hon callg d i y lh han

I - DLy Idn h a 1 cang c u n g ih hun

N6u hai tam g i i c c 6 hai canh tuung h g bang nhau tirng d6i mot nhung c i c g 6 c xen giDa khBng b i n g nhau thi canh t h ~ ? ba c6ng kh6ng bang nhau vh canh nho d6i d i t n vdi g6c ldn han II canh ldn hun

N t u hai tam g i i c c 6 hai canh tuung h g bang nhau tirng d6i tnot nhtrng canh thir ba kh6ng bang nhau thi g6c xen giira hai canh d 6 cGng kh6ng bang nhau vii g6c nho doi dign vdi canh 1dn han lii g 6 c l h han

V i du 15 C h h g minh r i n g duimg kinh di qua di6m chinh giira c6a mot cung thi di q u a trung didm c6a day cang cung ay Menh d& dHo c 6 dling kh8ng ? HHy n t u di&u kign dd menh dd d l o c i n g d6ng

Gitii

G i l sir d u h g kinh M N di qua didm chinh giDa M clia cung A B c i ~ a duirng trbn ( 0 )

X6t hai trubng h a p :

a) M l i d i h chinh giira cc6 cung nhb AB (h.22;) - - -

Vi M I i didm chinh giira cua cung AB n@n MA = MB suy ra MA = MB;do

d6 M tr&n d u h g trung t ~ c cha AB

OA = OB (bin kinh d u h g trbn (0)) do d 6 0 trCn dubng trung truc

c i a AB

Vgy OM l i d u h g trung truc c6a AB ntn OM di qua trung didm I ciia AB b) M l i didm chinh giira c6a cung lim AB (h.22b)

Vi c i c diim A v i B n8m trtn c i c nira d u h g trbn d u h g kinh MN n&n

c i c - cung - MA v i cung MB 18 c l c cung nho c ~ i a d t r h g trbn ( 0 ) m i

MA = MB n t n MA = MB, d o d6 M tr&n trung truc c6a AB Didm 0 trtn trung truc cha AB, vi th& dubng kinh MN 1s dubng trung truc cua AB ntn

MN c i t AB I3 I 18 trung di6m c3a AB

M&nh d & ddo khBng dung trong t r u h g hap AB is dubng kinh c i a duimg trbn (0) Ching han dubng kinh CD di qua trung didm 0 c6a dAy

AB nhung hai didrn C v i D kh6ng I i didm chinh giira c6a c l c nira dubng trbn dubng kinh AB

Didu ki&n dd m&nh d6 ddo ciing ddng 18 : Dtly AB kh6ng di qua tarn 0 M&nh d& d6 duuc phit bidu nhu sau : " D u h g kinh di qua trung didm c6a mat day cung kh6ng di qua tam dubng trbn thi di qua trung didm c i a day cang cung do"

121 Cho tam g i i c ctln ABC (AB = AC) Tr@n tia d6i c i a tia BA l&'y didm D

VC dubng trbn tAm 0 ngoai ti&p tam g i i c BCD - -

a) So s l n h s 6 d o c i c cung DB, BC v i 6?

b) K6 01, OH, OK l&n luqt vu6ng g6c v&i DC, DB, BC So sinh c5c doan 01, OH, OK

122 Cho tam g i l c ABC, AB < AC VE dubng trbn d u h g kinh BC c i t AB,

AC Idn luqt 6 E v i D ChCmg minh BD < CE

123 Cho d u h g trbn ( 0 ) day cung AB Goi C v l D 1s hai di6m tr&n AB sao cho AC = CD = DB G c tia OC, OD cit d u h g t r h ( 0 ) - - 1.9n l w I3 M v i PI

So sinh c i c cung AM, MN v i 6%

124 d u h g trbn Bigt Cho nira dubng trbn ( 0 ) dubng kinh AE = 2AB - - ; AD = 3 2 ; B, C, D lA ba didm trtn nifa a) C h h g minh ring AB = BC = CD ;

b) Chirng minh AC = BD ;

- A

C) C h h g minh ring c i c cung AD v l BC c6 chung didm chinh gisa M I

I d) Tlr g i i c ABCD 12 hinh gi ? Vi sao 3

125 Cho dubng trbn (0) duirng kinh AC v1 di6m B tren dubng trbn sao cho

= 60' Qua B k i day BD I AC, qua D k6 day DF//AC

, .A,

a) Tinh s 6 do c6c cung CD, AB, FD

b) Tim tigp tuygn c6a d u h g trbn (0) song song vdi AB ? I

G6c nei tigp 118 g6c c6 dinh nim tren d u h g trbn V i hai canh c k

d u h g trbn d6 #

Cung n l m ben trong g6c goi I l cung bj chin

SB do g6c nOi tigP bing nira s6 do cung bj chin

Trong met d u h g trbn :

- CXc g6c nQi ti&p c h g chan met cung hoac chin hai cung bang nhau

- Moi g6c nOi tigp chdn nira d u h g trbn d&u 18 g6c vu6ng

- Moi g6c nOi tigp (nht, hun hoac bing 9 c6 s 6 d o blng nira s 6 do c6a g6c b tam chng chin mot cung

Vi & 16 Cho tam g i i c can ABC (AB = AC) nOi ti6p d u h g trbn (0)

Tia phan g i i c cha g6c B vP g6c C c i t d u h g trbn b D vP E

a) So s i n h hai tam g i i c ACE v1 ABD ;

b) Goi I I i giao di6m c6a BD v2 CE Tir giic ADIE I 1 hinh gi ? Vi sao ?

Tam - giBc 'ABC can A

suy ra Lai ABD c6 ADB = - = ACE e (hai g6c nai ti* cbng

~ i n h 23

- - - - than cung AB), AEC = B (hai g6c noi tie'p chng chan cung AC) nen

ABD = ACE n&n AD = AE, suy ra AD = AE

Hinh binh hinh ADIE c 6 hai canh kk AD = AE n&n l i hinh thoi

Vi dl;, 17 Cho hai d u h g trbn (0) va ( 0 ' ) cit nhau b A v2 B ( 0 v i 0' thuoc hainira mat phing bit AB) Qua A k i cit tuygn cht duitng trbn ( 0 ) a C, c i t

d u h g trbn (0') b D X i c djnh vi tri c i t tuye'n CD d6 CD c6 do d i i I& nhgt

A v i B nen 00' 1 AB tai I l i trung

I Tam g i i c AOB can 13 0 , c 6 0 1 1i duitng cao n&n 1 i phin g i i c cha g6c