documents.tips huong trong hinh hoc phang nguyen minh hapdf

Nguyn Minh H

NH XUT BN DN TR

Mc lc

Li gii thiu iii

Li ni u v

Cc k hiu vii

Chng 1. Hng ca on thng 1

1. Hnh thang v hnh bnh hnh 1

2. on thng 2

3. on thng nh hng, hng v phng ca n 43.1. Cc nh ngha 43.2. Cc nh l 103.3. Hng ca on thng nh hng 17

4. Vect, hng v phng ca n 184.1. Cc nh ngha 184.2. Cc nh l 194.3. Hng v phng ca vect 20

5. Hng v phng ca tia 215.1. Cc nh ngha 215.2. Cc nh l 225.3. Hng v phng ca tia 26

6. Hng hn tp, phng hn tp, ng thng nh hng 276.1. Hng hn tp, phng hn tp 276.2. ng thng nh hng 29

7. di i s ca on thng nh hng 31

Chng 2. Hng ca gc 37

8. Gc gia hai tia 37

9. Gc nh hng gia hai tia v cc vn c lin quan 449.1. Gc nh hng gia hai tia 44

i

ii Mc lc

9.2. C s, tia c s ca gc nh hng gia hai tia-khc btc cng nh 45

9.3. S khng trng lp, s trng lp ca hai gc nh hnggia hai tia-khc bt c cng nh 46

9.4. Ngun v ct tuyn ca hai gc nh hng gia haitia-khc bt c cng nh 47

9.5. Cc nh l v ct tuyn ca hai gc nh hng gia haitia-khc bt cng nh 51

10. S cng hng, s ngc hng ca hai gc nh hnggia hai tia 63

10.1. Hai gc nh hng gia hai tia c cng nh 6310.2. Hai gc nh hng gia hai tia bt k 6810.3. Hng ca gc nh hng gia hai tia, mt phng

nh hng 7610.4. Hng ca tam gic v hng ca a gic li 77

11. S o ca gc nh hng, gc lng gic gia hai tia 8111.1. S o ca gc nh hng gia hai tia 8111.2. Gc lng gic gia hai tia 82

12. Gc, gc nh hng, gc lng gic gia hai vect 8612.1. Gc gia hai vect 8612.2. Gc nh hng gia hai vect 8712.3. Gc lng gic gia hai vect 90

13. Cung, cung nh hng, cung lng gic 9113.1. Cung 9113.2. Cung nh hng 9213.3. Cung lng gic 94

14. Gc, gc nh hng, gc lng gic gia hai ng thng 9514.1. Gc gia hai ng thng 9514.2. Gc nh hng gia hai ng thng 9714.3. Gc lng gic gia hai ng thng 101

15. Mt vi kt qu c bn 105

Ti liu tham kho 109

Tra cu theo vn 111

Li gii thiu

Cng bn c,Thu cn l hc sinh ph thng, khi hc bi gc lng gic ti cm

thy c g bt n nhng khng hiu v sao mnh li c cm gic .Sau ny, sng bng ngh dy Ton v lm ton, ti mi hiu rng cing h chnh l nguyn nhn ca s bt n , khi nim gc lnggic c nh ngha thng qua ci ng h nhng ci ng h li khngphi l khi nim ca hnh hc phng. Trong h tin Hilbert cahnh hc Euclid phng, gi tt l hnh hc phng, mi khi nim phic nh ngha thng qua hai khi nim c bn: im, ng thng vba quan h c bn: lin thuc, nm gia, ton ng.

Mi m vi cng vic ring ca mnh, ti khng h ngh rng lic mt ngi quan tm n vic nh ngha gc lng gic m khngs dng ci ng h, ni theo cch ca nhng ngi lm ton chuynnghip, quan tm ti vn xy dng l thuyt v hng trong hnhhc phng.

Cm trong tay bn tho hn mt trm trang cun sch Hngtrong hnh hc phng, hn mt trm trang m vit trong hn minm tri, ti thc s bt ng v ci tnh yu m thm v bn b m tcgi ca n, TS Nguyn Minh H dnh cho Ton hc.

Vi nhng g m ti bit v TS Nguyn Minh H, vi cch t vn rt hp l ca Hng trong hnh hc phng, chc rng cun schny l mt ti liu rt ng c cho bt k ai quan tm ti hnh hcphng, c bit l sinh vin khoa Ton ca cc trng i hc s phmv Cao ng s phm.

Hy c Hng trong hnh hc phng xem ci cch m TSNguyn Minh H vt ci ng h ra khi hnh hc phng.

GS. TSKH Nguyn Vn Khu

iii

Li ni uTi xy cn nh nh ca ti

trong to nh ln ca Hilbert v Euclid

Hng l khi nim quan trng ca hnh hc. Tuy nhin, t thi Euclidcho ti trc thi ca Descartes hng khng c coi l khi nim cahnh hc. T khi c phng php to ca Descartes tnh trng trn phn no c gii quyt, bng cc khi nim ma trn v nh thchng tr thnh khi nim ca hnh hc. Ch rng phn no cgii quyt ch khng phi hon ton c gii quyt, khi khng cphng php to ngi ta vn ch c th ni ti hng di dng mt. V vy nhng vn lin quan ti hng thng b n trnh, trongton b tc phm C s hnh hc ca Hilbert [1] khng c dng nodnh cho khi nim hng.

Khng c khi nim hng, khng th trnh by mt cch cht chnhiu vn ca hnh hc (gc lng gic, vect, l thuyt bin hnh,. . . ). Khng c khi nim hng, ta gp rt nhiu kh khn trong hctp, ging dy v nghin cu hnh hc. Khng c khi nim hng, hnhhc phng - mt trong nhng ngnh khoa hc c xa nht ca nhn loi- tng nh khng cn iu g ng bn sau khi Hilbert vit tc phmC s hnh hc cho n ngy hm nay vn cha hon chnh.

V sao li c phi n trnh? Liu c th ni ti khi nim hngm khng cn s dng phng php to hay khng? Nhiu nmnay nhng cu hi ny thi thc ti hng ti mc tiu: xy dngl thuyt v hng, trc ht l trong hnh hc phng, khng s dngphng php to , tt cho vic lm ton. Gi y l thuyt ny c xy dng xong. Cun sch Hng trong hnh hc phng mbn ang c trong tay cha ng ton b l thuyt , n bao gm haichng: Chng I-Hng ca on thng; Chng II-Hng ca gc.

Ti dnh li cm n ti bn Nguyn Duy Khnh, ngi c nhiung gp trong vic trnh by v bin tp cun sch.

Ti rt mong nhn c nhng nhn xt qu gi t c gi.

Nguyn Minh H

v

Cc k hiu

A =B: cc im A, B trng nhau.

A 6=B: cc im A, B khc nhau.

A,B / XY : hai im A,B cng thuc mt na mt phng b XY .

A / XY / B: hai im A,B thuc hai na mt phng khc nhau b XY .

AB: on thng c hai u mt l cc im A,B.

AB: di on thng AB (nu khng c g nhm ln).

AB: ng thng i qua hai im phn bit A,B (nu khng c g nhmln).

a b: cc ng thng a,b trng nhau.

a 6 b: cc ng thng a,b khng trng nhau.

a b: cc ng thng a,b song song.

a b: cc ng thng a,b hoc song song hoc trng nhau.

a b: cc ng thng a,b vung gc.

vii

viii Cc k hiu

a 6 b: cc ng thng a,b khng vung gc.

ab=O: cc ng thng a, b ct nhau ti im O.#

AB: on thng nh hng c u mt u l im A, u mt cui lim B.

#0 : on thng nh hng-khng.

#

AB # CD: cc on thng nh hng # AB, # CD cng hng.#

AB # CD: cc on thng nh hng # AB, # CD ngc hng.#

AB # CD: cc on thng nh hng # AB, CD cng phng.#

AB= # CD: cc on thng nh hng # AB, # CD bng nhau.#

AB 6= # CD: cc on thng nh hng # AB, # CD khc nhau.#

AB: vect cha on thng nh hng#

AB (nu khng c g nhm ln).

[#

AB]: vect cha on thng nh hng#

AB.

#0 : vect-khng (nu khng c g nhm ln).

[#0 ]: vect-khng.

#a #b : cc vect #a , #b cng hng.#a #b : cc vect #a , #b ngc hng .#a #b : cc vect #a , #b cng phng.#a #b : cc vect #a , #b vung gc.#a = #b : cc vect #a , #b bng nhau.#a 6= #b : cc vect #a , #b khc nhau.

Cc k hiu ix

#

AB: tia c gc l im A v i qua im B (nu khng c g nhm ln).

Ix Jy: cc tia Ix,Jy cng hng.

Ix Jy: cc tia Ix,Jy ngc hng.

Ix Jy: cc tia Ix,Jy trng nhau.

Ix Jy: cc tia Ix,Jy cng phng.

Ix Jy: cc tia Ix,Jy vung gc.

Ix Jy: tia Ix thuc tia Jy.

Ox: tia i ca tia Ox.

xx: ng thng cha hai tia Ox v Ox.

yx: ng thng cha hai tia i nhau Ox v Oy.

xOy: gc gia hai tia c cc cnh l cc tia Ox,Oy.AOB: gc gia hai tia c cc cnh l cc tia # OA, # OB.AOy: gc gia hai tia c cc cnh l cc tia # OA,Oy.xOB: gc gia hai tia c cc cnh l cc tia Ox, # OB.(Ox,Oy): gc nh hng gia hai tia c cnh u l tia Ox, cnh cui ltia Oy.

(#

OA,#

OB): gc nh hng gia hai tia c cnh u l tia#

OA, cnh cuil tia

#

OB.

(#

OA,Oy): gc nh hng gia hai tia c cnh u l tia#

OA, cnh cuil tia Oy.

(Ox,#

OB) gc nh hng gia hai tia c cnh u l tia Ox, cnh cui ltia

#

OB.

x Cc k hiu

(Ix, I y) (Jz,Jt): cc gc nh hng gia hai tia (Ix, I y), (Jz,Jt) cnghng.

(Ix, I y) (Jz,Jt): cc gc nh hng gia hai tia (Ix, I y), (Jz,Jt) ngchng.

4ABC: tam gic ABC.

4ABC 4XYZ: cc tam gic ABC, XYZ cng hng.

4ABC 4XYZ: cc tam gic ABC, XYZ ngc hng.

(Ox,Oy)k: gc lng gic gia hai tia c gc nh hng gia hai tiasinh l (Ox,Oy) v c chu k l k.

#a , #b : gc gia hai vect c cc cnh l cc vect #a , #b .

(#a ,#

b ): gc nh hng gia hai vect c cnh u l vect #a , cnh cuil vect

#

b .

(#a ,#

b ) (#c , #d ): cc gc nh hng gia hai vect (#a , #b ), (#c , #d ) cnghng.

(#a ,#

b ) (#c , #d ): gc nh hng gia hai vect (#a , #b ), (#c , #d ) ngc hng.

(#a ,#

b )k: gc lng gic gia hai vect c gc nh hng gia hai vectsinh l (#a ,

#

b ) v c chu k l k.

AB: cung c hai u mt l cc im A,B.AB: cung nh hng c umt u l im A, umt cui l im B.AB CD: cc cung nh hngAB,CD cng hng.AB CD: cc cung nh hngAB,CD ngc hng.ABk: cung lng gic c cung nh hng sinh l AB v c chu k lk.

a,b: gc gia hai ng thng c cc cnh l cc ng thng a,b.

Cc k hiu xi

(a,b): gc nh hng gia hai ng thng c cnh u l ng thnga, cnh cui l ng thng b.

(a,b) (c,d): gc nh hng gia hai ng thng (a,b), (c,d) cnghng.

(a,b) (c,d): gc nh hng gia hai ng thng (a,b), (c,d) ngchng.

(a,b)k: gc lng gic gia hai ng thng c gc nh hng giahai ng thng sinh l (a,b) v c chu k l k.

: Kt thc mt php chng minh.

Chng 1

Hng ca onthng

1. Hnh thang v hnh bnh hnh

Theo quan nim thng thng, hnh thang v hnh bnh hnh l haikhi nim khc nhau v c nh ngha nh sau.

nh ngha 1. Hnh thang l t gic li c mt b hai cnh i thuchai ng thng song song v b hai cnh i cn li thuc hai ngthng khng song song.

nh ngha 2. Hnh bnh hnh l t gic li m mi b hai cnh icng thuc hai ng thng song song.

nh ngha 1 c v r rng nhng li khng ph hp vi tinh thnca l thuyt tp hp. Do n khng thun tin cho vic lm ton. Vvy, gn y, trong nhiu ti liu ngi ta nh ngha hnh thang nhsau.

nh ngha 3. Hnh thang l t gic li c mt b hai cnh i thuchai ng thng song song.

Trong nh ngha 3, b hai cnh i cn li ca hnh thang c ththuc hai ng thng hoc song song hoc khng song song. Do hnh bnh hnh l mt hnh thang c bit.

1

2 1. Hng ca on thng

nh ngha 3 ph hp vi tinh thn ca l thuyt tp hp. V vyn thun tin cho vic lm ton. Do n c coi l nh ngha chnhthng v chnh thc c s dng trong cun sch ny.

Mi mt trong hai cnh thuc b hai cnh i thuc hai ng thngsong song ca hnh thang c gi l cnh y ca n. Mi mt tronghai cnh thuc b hai cnh i cn li ca hnh thang c gi l cnhbn ca n. Do , i vi hnh bnh hnh, mt hnh thang c bit, tac hai cch quan nim v cnh y v cnh bn. C th, vi hnh bnhhnh ABCD, nu coi AB,CD l cnh y th AD,CB l cnh bn, nucoi AD,CB l cnh y th AB,CD l cnh bn.

Theo cch k hiu thng thng, mt t gic li K c bn nh lX ,Y ,Z,T v bn cnh l XY ,YZ,ZT,TX c k hiu l ABCD, trong (A,B,C,D) l mt hon v ca (X ,Y ,Z,T) v (AB,BC,CD,DA) l mthon v ca (XY ,YZ,ZT,TX ). Do K c k hiu bi mt trong tmcch sau: XYZT,YZTX ,ZTXY ,TXYZ,XTZY ,TZY X ,ZY XT,Y XTZ.

Theo thi quen, cch k hiu trn cng c dng k hiu hnhthang K c bn nh l X ,Y ,Z,T v bn cnh l XY ,YZ,ZT,TX . Tuynhin, vi cch k hiu ny ta khng th bit c cnh no trong bncnh XY ,YZ,ZT,TX l cnh y caK . l nguyn nhn ca nhiubt li trong vic lm ton. V vy, trong cun sch ny hnh thang Kc bn nh l X ,Y ,Z,T, bn cnh l XY ,YZ,ZT,TX v hai cnh yl XY ,ZT c k hiu l ABCD, trong (A,B,C,D) l mt hon vca (X ,Y ,Z,T), (AB,BC,CD,DA) l mt hon v ca (XY ,YZ,ZT,TX )v (AB,CD) l mt hon v ca (XY ,ZT). Do K c k hiu bi mttrong bn cch sau: XYZT,ZTXY ,Y XTZ,TZY X .

Ch 4. 1) Nu mt trong bn t gic li ABCD,CDAB,BADC,DCBAl hnh thang th c bn cng l hnh thang.

2) Numt trong tm t gic li ABCD,BCDA,CDAB,DABC,ADCB,DCBA,CBAD,BADC l hnh bnh hnh th c tm cng l hnh bnhhnh.

2. on thng

Trong mc ny, mt s kin thc c bn v on thng c nhc li.

nh ngha 5. B khng phn bit th t gm hai im khc nhauA,B c gi l on thng, hoc k hiu l AB hoc k hiu l BA.

nh ngha 6. Giao ca cc tia AB, BA c gi l min trong caon thng AB.

2. on thng 3

nh ngha 7. B khng phn bit th t gm hai im trng nhauA,B cng c gi l on thng (on thng-khng, khi cn nhnmnh), k hiu bi mt trong cc cch sau: AB,BA,AA,BB.

nh ngha 8. Min trong ca on thng-khng l tp hp rng.

Cc im A,B c gi l u mt ca on thng AB.Vi s xut hin ca khi nim on thng-khng, thut ng on

thng mang mt ngha mi: on thng c th l on thng-khckhng (hai u mt khc nhau) v cng c th l on thng-khng(hai u mt trng nhau). V vy, trnh nhm ln, trong cc vn lin quan ti khi nim on thng, khi cn thit ta s nhn mnh ccthut ng: on thng-khc khng, on thng-khng.

Ch 9. 1) Gc O ca tia Ox khng thuc tia Ox.2) Hnh gm tia Ox v im O c gi l tia Ox m rng.3) Cc u mt ca on thng AB khng thuc min trong ca

on thng AB.

Nu ta qui c mt on thng no c di bng 1 th i vimi on thng tn ti duy nht mt s thc dng biu th di caon thng .

Nu khng c g nhm ln th di on thng AB c k hiun gin l AB.

Ch 10. AB= 0 khi v ch khi AB l on thng-khng.nh ngha 11. Hai on thng AB,CD c gi l bng nhau nuchng c di bng nhau.

biu th hai on thng AB,CD bng nhau, ta vit AB=CD.nh l 12. Nu im C hoc thuc min trong hoc trng vi mt tronghai u mt ca on thng AB th AB= AC+CB (h thc Chasles choon thng).

Trong nh l 12, on thng AB c th l on thng-khng.

Ch 13. 1) Thay cho cch ni im C thuc min trong ca onthng AB ta cn c cc cch ni n gin hn: im C thuc on thngAB, im C nm trong on thng AB, im C nm gia cc im A,B.

2) Nu im C thuc ng thng AB, khng nm trong on thngAB, khc cc im A,B th ta ni im C nm ngoi on thng AB.

4 1. Hng ca on thng

nh ngha 14. Hnh thang c ng mt cnh y l on thng-khng c gi l hnh thang-khng.

h.1 h.2

A

B C B

A

C =D

Tam gic ABC khng phi l hnh thang-khng nhng t gic ABCDvi C = D l hnh thang-khng c mt cnh y l on thng-khckhng AB v mt cnh y l on thng-khng CD (h.1, h.2).

Vi s xut hin ca khi nim hnh thang-khng, thut ng hnhthang mang mt ngha mi: hnh thang c th l hnh thang-khckhng (hai cnh y l nhng on thng-khc khng) v cng c thl hnh thang-khng (c ng mt cnh y l on thng-khng). Vvy, trnh nhm ln, trong cc vn lin quan ti khi nim hnhthang, khi cn thit ta s nhn mnh cc thut ng: hnh thang-khckhng, hnh thang-khng.

3. on thng nh hng, hng v phng ca n

Trong mc 2, khi nh ngha on thng ta khng phn bit th t haiu mt ca n. Trong mc ny, ta lm quen vi mt khi nim mi:on thng m th t hai u mt ca n phn bit.

3.1. Cc nh ngha.

nh ngha 15. B c phn bit th t gm hai im (A,B) c gi lon thng nh hng, k hiu l

#

AB.

Khi cc im A,B trng nhau, on thng nh hng#

AB c gil on thng nh hng-khng, cn k hiu bi mt trong cc cchsau:

#

BA,#

AA, # BB.Cc im A,B theo th t c gi l u mt u, u mt cui

ca on thng nh hng#

AB.Khi cn thit, thay cho thut ng on thng nh hng ta c th

nhn mnh bng cc thut ng: on thng nh hng-khc khng

3. on thng nh hng, hng v phng ca n 5

(hai u mt khc nhau); on thng nh hng-khng (hai u mttrng nhau).

nh ngha 16. di ca on thng nh hng # AB l di caon thng AB.

di ca on thng nh hng#

AB c k hiu l |# AB|. Vy|# AB| = AB.Ch 17. |# AB| = 0 khi v ch khi # AB l on thng nh hng-khng.

on thng AB c gi l on thng sinh ca on thng nhhng

#

AB. Vy di ca on thng nh hng l di ca onthng sinh ca on thng nh hng .

B sau y khng ch gip ta chng minh b 20 m cn c vaitr quan trng trong nhiu tnh hung khc.

B 18. (B hnh thang) Nu AB CD th1) Hoc A,D / BC hoc A / BC / D.2) A,D / BC khi v ch khi ABCD l hnh thang.3) A / BC / D khi v ch khi ABDC l hnh thang.

Chng minh. 1) Hin nhin.2) iu kin cn. V AB CD nn A,B / CD v C,D / AB (1).Gi () l giao ca na mt phng b CD cha A,B v na mt

phng b AB cha C,D (h.3).

()

h.3

E

A B

D C

C hai trng hp cn xem xt.Trng hp 1. AD, BC song song. Hin nhin B,C / AD.Trng hp 2. AD, BC khng song song. Gi E l giao im ca AD

v BC. V A,D / BC nn E khng thuc on thng AD. Kt hp vi

6 1. Hng ca on thng

on thng AD thuc (), suy ra E khng thuc (). T , ch rngon thng BC thuc (), suy ra E khng thuc on thng BC. iu c ngha l B,C / AD.

Tm li, trong c hai trng hp ta u c B,C / AD (2).T (1) v (2), ch rng A,D / BC, suy ra ABCD l t gic li. Kt

hp vi AB CD, suy ra ABCD l hnh thang.iu kin . V ABCD l hnh thang nn ABCD l t gic li. Do

A,D / BC.3) iu kin cn. V AB CD nn A,B / CD v C,D / AB (1).Gi () l giao ca na mt phng b CD cha A, B v na mt

phng b AB cha C, D.V A / BC / D nn AD v BC ct nhau. Gi F l giao im ca AD v

BC (h.4).

()

h.4

A B

DC

F

V A / BC / D nn F thuc on thng AD. Do A,F / BD vD,F / AC (2).

V F thuc on thng AD nn F thuc (). T ch rng onthng BC thuc (), suy ra F thuc on thng BC. Do C,F / BD vB,F / AC (3).

T (2) v (3) suy ra A,C / BD v B,D / AC (4).T (1) v (4) suy ra ABDC l t gic li. Kt hp vi AB CD, suy

ra ABDC l hnh thang.iu kin . V ABDC l hnh thang nn ABDC l t gic li. Do

A / BC / D.

Ch 19. Na mt phng b a v ng thng a khng c im chung.

B 20. (B ba hnh thang) Nu ba ng thng AB,CD,XY imt khng trng nhau v ABY X ,DCY X l hnh thang th ABCD cngl hnh thang.


Chng minh. Gi Z l giao im ca XY v BC. C bn trng hpcn xem xt.

Trng hp 1. Z =Y (h.5).

h.5

A B

CD

XY = Z

V Z =Y nn Y thuc BC. Do BY BC CY . V ABY X v DCY Xl hnh thang nn, theo b 18, A,X / BY v D,X / CY . Vy A,D / BC(1).

V ABY X v DCY X l hnh thang nn AB XY v CD XY . Kthp vi AB 6CD, ta c AB CD (2).

T (1) v (2), theo b 18, suy ra ABCD l hnh thang.Trng hp 2. Z = X (h.6).

h.6

A B

CD

X = Z Y

V Z = X nn X thuc BC. Do BX BC CX . V ABY X v DCY Xl hnh thang nn, theo b 18, A / BX / Y v D / CX / Y . Vy A,D / BC(1).

V ABY X v DCY X l hnh thang nn AB XY v CD XY . Kthp vi AB 6CD, ta c AB CD (2).

T (1) v (2), theo b 18, suy ra ABCD l hnh thang.Trng hp 3. Z thuc tia # XY v Z 6=Y (h.7).

8 1. Hng ca on thng

h.7

A B

CD

X YZ

V ABY X v DCY X l hnh thang nn, theo ch 4, BAXY vCDXY cng l hnh thang. Do , theo b 18, B,Y / AX v C,Y / DX .T , ch rng Z thuc tia # XY , suy ra B,Z / AX v C,Z / DX . Kthp vi AB XZ v CD XZ, theo b 18, suy ra BAXZ v CDXZ lhnh thang. Vy, li theo ch 4, ABZX v DCZX cng l hnh thang.Do , theo trng hp 1, ABCD l hnh thang.

Trng hp 4. Z thuc tia i ca tia # XY (h.8).

h.8

A B

CD

Z X Y

V ABY X v DCY X l hnh thang nn, theo b 18, A,X / BY vD,X / CY . T , ch rng Z thuc tia i ca tia # XY , suy ra A,Z / BYv D,Z / CY . Kt hp vi AB YZ v DC YZ, li theo b 18, suy raABYZ v DCYZ cng l hnh thang. Do , theo trng hp 2, ABCDl hnh thang.

B 20 khng nh s hp l ca cc nh ngha 21, 22.

nh ngha 21. Hai on thng nh hng # AB, # CD c gi l cnghng nu tn ti on thng-khc khng XY sao cho cc t gic ABY Xv CDY X l hnh thang (c th l hnh thang-khng) (h.9, h.10, h.11,h.12, h.13, h.14).


h.9 h.10

h.11 h.12

h.13 h.14

A B

YX

C D

X Y

A BC D

A B

YX

C =D

X Y

A BC =D

A =B

YX

C =D

X Y

A =B=C =D

biu th#

AB,#

CD cng hng, ta vit#

AB # CD.nh ngha 22. Hai on thng nh hng # AB, # CD c gi l ngchng nu tn ti on thng-khc khng XY sao cho cc t gic ABY Xv CDXY l hnh thang (c th l hnh thang-khng) (h.15, h.16, h.17,h.18, h.19, h.20).

h.15 h.16

A B

YX

D C

X Y

A BD C

10 1. Hng ca on thng

h.17 h.18

h.19 h.20

A B

YX

C =D

X Y

A BC =D

A =B

YX

C =D

X Y

A =B=C =D

biu th#

AB,#

CD ngc hng, ta vit#

AB # CD.

nh ngha 23. Hai on thng nh hng # AB, # CD c gi l cngphng nu chng hoc cng hng hoc ngc hng.

biu th#

AB,#

CD cng phng, ta vit#

AB # CD.

nh ngha 24. Hai on thng nh hng # AB, # CD c gi l bngnhau nu chng cng hng v c di bng nhau.

biu th#

AB,#

CD bng nhau (khc nhau), ta vit#

AB = # CD (# AB 6=#

CD).

3.2. Cc nh l.

nh l 25. on thng nh hng-khng cng hng vi mi onthng nh hng.

Chng minh. Gi s # AA l on thng nh hng-khng v # BC lon thng nh hng bt k. C hai trng hp cn xem xt.

Trng hp 1. B=C (h.21).Dng ng thng XY khng i qua A = A v B = C. V AAY X v

BCY X cng l hnh thang-khng nn#

AA # BC.


h.21 h.22

A = A

X Y

B=C

X Y

A = A

B C

Trng hp 2. B 6=C (h.22).Dng ng thng XY khng i qua A = A v song song vi BC sao

cho BCY X l hnh thang. V AAY X l hnh thang-khng v BCY X lhnh thang nn

#

AA # BC. nh l 26. on thng nh hng-khng ngc hng vi mi onthng nh hng.

Chng minh. Gi s # AA l on thng nh hng-khng v # BC lon thng nh hng bt k. C hai trng hp cn xem xt.

Trng hp 1. B=C (h.23).Dng ng thng XY khng i qua A v B=C. V AAY X v BCXY

cng l hnh thang-khng nn#

AA # BC.

h.23 h.24

A = A

X Y

B=C

X Y

A = A

C B

Trng hp 2. B 6=C (h.24).Dng ng thng XY khng i qua A = A v song song vi BC sao

cho BCXY l hnh thang. V AAY X l hnh thang-khng v BCXY lhnh thang nn

#

AA # BC. nh l 27. on thng nh hng-khng cng phng vi mi onthng nh hng.


nh l 27 l h qu trc tip ca cc nh l 25, 26.

nh l 28. Cc on thng nh hng-khng bng nhau

Chng minh. Theo nh l 25, cc on thng nh hng-khng cnghng. T , theo ch 10, suy ra cc on thng nh hng-khngbng nhau.

Ch 29. Bi nh l 28, cc on thng nh hng-khng cng ck hiu l #0 . Vy, vi mi im A,

#

AA = #0 .nh l 30. Vi hai ng thng AB v CD khng trng nhau, ta cABCD l hnh thang khi v ch khi

#

AB # DC.

Chng minh. iu kin cn. Dng ng thng song song vi ccng thng AB,CD v ct cc on thng AD, BC tng ng ti X ,Y(h.25).

h.25

A B

D C

X Y

V ABCD l hnh thang nn, theo b 18, A,D / BC. T , ch rng X ,Y theo th t thuc cc on thng AD, BC, suy ra A,X / BYv D,X / CY . Kt hp vi AB Y X v DC Y X , li theo b 18, suy raABY X v DCY X l hnh thang. Vy

#

AB # DC.iu kin . V

#

AB # DC nn tn ti on thng-khc khng XYsao cho ABY X v DCY X l hnh thang. T , ch rng cc ngthng AB,CD,XY i mt khc nhau, theo b 20, suy ra ABCD lhnh thang.

nh l 31. Vi hai ng thng AB,CD, ta c AB CD khi v ch khi#

AB # CD.

Chng minh. iu kin cn. Dng ng thng song song vi ccng thng AB,CD. Trn ly cc im X ,Y sao cho ABY X l hnhthang (h.26, h.27, h.28, h.29).


Theo b 18, hoc CDY X hoc CDXY l hnh thang. Do , theonh l 30, hoc

#

AB # CD hoc # AB # CD. Vy # AB # CD.

h.26

h.27

h.28

h.29

X

A B

Y

C D

X Y

A BC D

X

A B

Y

D C

X Y

A BD C

iu kin . V#

AB # CD nn hoc # AB # CD hoc # AB # CD. Do tn ti on thng-khc khng XY sao cho hoc ABY X v CDY X lhnh thang hoc ABY X v CDXY l hnh thang. iu c ngha l tnti ng thng XY sao cho AB XY v CD XY . Vy AB CD. nh l 32. Vi hai im A,B, ta c

1)#

AB # AB.2)

#

AB # BA.Chng minh. Nh cc nh l 25, 26, ta ch cn chng minh nh l32 trong trng hp A 6= B. Dng on thng-khc khng XY sao choABY X l hnh thang (h.30).


h.30

A B

YX

1) V ABY X l hnh thang nn ABY X v ABY X l hnh thang. Do

#

AB # AB.2) V ABY X l hnh thang nn, theo ch 4, ABY X v BAXY l

hnh thang. Do #

AB # BA.

nh l 33. Vi hai on thng nh hng # AB, # CD, ta c # AB # CD khiv ch khi

#

AB # DC.

Chng minh. Nh cc nh l 25, 26, ta ch cn chng minh nh l 33trong trng hp

#

AB 6= #0 v # CD 6= #0 .

h.31

A B

DC

X Y

Cc iu kin sau tng ng (h.31).

1)#

AB # CD.2) Tn ti on thng-khc khng XY sao cho ABY X v CDY X l

hnh thang.3) Tn ti on thng-khc khng XY sao cho ABY X v DCXY l

hnh thang.4)

#

AB # DC.Ch , hin nhin 1 2; theo ch 4, 2 3; hin nhin, 3 4.


nh l 34. Nu ba im A,B,C i mt khc nhau th

1) C nm ngoi on thng AB khi v ch khi#

CA # CB.2) C nm trong on thng AB khi v ch khi

#

CA # CB.

Chng minh. 1) iu kin cn. Khng mt tnh tng qut gi s Cthuc tia i ca tia

#

AB. Dng on thng-khc khng XY sao choCBY X l hnh thang (h.32).

V CBY X l hnh thang nn, theo ch 4, BCXY cng l hnhthang. Do , theo b 18, B,Y / CX . V C thuc tia i ca tia

#

ABnn A,B / CX . Vy A,Y / CX . T , ch rng CA XY , li theo b 18, suy ra ACXY l hnh thang. Li theo ch 4, CAXY cng l hnhthang.

h.32

C B

YX

A

Tm li CBY X v CAY X l hnh thang. iu c ngha l#

CA #

CB.iu kin . V

#

CA # CB nn tn ti on thng-khc khng XYsao cho CAY X v CBY X l hnh thang (h.32).

Do , theo ch 4, ACXY v BCXY cng l hnh thang. T ,theo b 18, suy ra A,Y / CX v B,Y / CX . Vy A,B / CX . Kt hp viCA XY CB, suy ra C thuc ng thng AB nhng khng nm trongon thng AB. Ni cch khc, theo ch 13, C nm ngoi on thngAB.

2) iu kin cn. Dng on thng-khc khng XY sao cho CAXYl hnh thang (h.33).

V CAXY l hnh thang nn, theo ch 4, ACY X cng l hnhthang. Do , theo b 18, A / CX / Y . V C thuc on thng AB nnA / CX / B. Vy B,Y / CX . Kt hp vi CB XY , li theo b 18, suy raBCXY l hnh thang. Li theo ch 4, CBY X cng l hnh thang.


h.33

A B

YX

C

Tm li CAXY v CBY X cng l hnh thang. iu c ngha l#

CA # CB.iu kin . V

#

CA # CB nn tn ti on thng-khc khng XYsao cho CAXY v CBY X l hnh thang (h.33).

Do , theo ch 4, ACY X v BCXY cng l hnh thang. T ,theo b 18, suy ra A / CX / Y v B,Y / CX . Vy A / CX / B. Kt hpvi CA XY CB, suy ra C thuc on thng AB. nh l 35. Vi ba on thng nh hng-khc khng # AB, # CD, # EF, tac

1) Nu#

AB # CD; # CD # EF th # AB # EF.2) Nu

#

AB # CD; # CD # EF th # AB # EF.3) Nu

#

AB # CD; # CD # EF th # AB # EF.

Chng minh. 1) V # AB # CD; # CD # EF nn # AB # CD; # CD # EF. Do ,theo nh l 31, cc ng thng AB,CD,EF i mt hoc song songhoc trng nhau.

Dng ng thng XY song song vi cc ng thng AB,CD,EF(h.34).

h.34A B

YX

C D

FE


Theo b 18, hoc ABY X l hnh thang hoc ABXY l hnh thang,hoc CDY X l hnh thang hoc CDXY l hnh thang, hoc EFY X lhnh thang hoc EFXY l hnh thang.

Khng mt tnh tng qut gi s ABY X l hnh thang (1).Nu CDXY l hnh thang th

#

AB # CD, mu thun. Vy CDY Xl hnh thang. Do , nu EFXY l hnh thang th

#

CD # EF, li muthun. Vy EFY X l hnh thang (2).

T (1) v (2), suy ra#

AB # EF.2) V

#

AB # CD; # CD # EF nn, theo nh l 33, # AB # CD; # CD # FE.Do , theo phn 1,

#

AB # FE. Vy, li theo nh l 33, # AB # EF.3) V

#

AB # CD; # CD # EF nn, theo nh l 33, # AB # DC; # DC # EF.Do , theo phn 1,

#

AB # EF.

3.3. Hng ca on thng nh hng.Theo cc nh l 32, 35, d dng thy rng trong tp hp cc on thngnh hng-khc khng quan h cng hng l quan h tng ng.

nh ngha 36. Mi lp tng ng sinh ra bi quan h cng hngtrong tp hp cc on thng nh hng-khc khng c gi l hngca on thng nh hng.

Hng ca on thng nh hng cha on thng nh hng#

ABc gi n gin l hng ca on thng nh hng

#

AB.Thut ng hng ca on thng nh hng gii thch phn no

ngha ca thut ng cng hng trong nh ngha 21. C th, hai onthng nh hng-khc khng c gi l cng hng nu chng cngthuc mt hng ca on thng nh hng.

Nu#

AB v#

CD l hai on thng nh hng-khc khng ngchng th, theo nh l 35, mi on thng nh hng thuc hng caon thng nh hng

#

AB ngc hng vi mi on thng nh hngthuc hng ca on thng nh hng

#

CD. Nhn xt ny khng nhs hp l ca nh ngha sau.

nh ngha 37. Hai hng ca on thng nh hng c gi lngc nhau nu mi on thng nh hng thuc hng ca onthng nh hng ny v mi mt on thng nh hng thuc hngca on thng nh hng kia ngc hng.

Theo nh l 31, d dng thy rng trong tp hp cc on thngnh hng-khc khng quan h cng phng cng l quan h tngng.


nh ngha 38. Mi lp tng ng sinh ra bi quan h cng phngtrong tp hp cc on thng nh hng-khc khng c gi l phngca on thng nh hng.

Phng ca on thng nh hng cha on thng nh hng#

AB c gi n gin l phng ca on thng nh hng#

AB.Thut ng phng ca on thng nh hng gii thch phn no

ngha ca thut ng cng phng trong nh ngha 23. C th, haion thng nh hng-khc khng c gi l cng phng nu chngcng thuc mt phng ca on thng nh hng.

Mi phng ca on thng nh hng cha hai hng ca onthng nh hng v l hai hng ca on thng nh hng ngcnhau.

4. Vect, hng v phng ca n

S dng cc khi nim: hai on thng nh hng cng hng, haion thng nh hng ngc hng, d dng i n cc khi nim:hng ca vect, phng ca vect.

4.1. Cc nh ngha.Theo cc nh l 32, 35, ch rng trong tp hp cc on thng quanh bng nhau l quan h tng ng, d dng thy rng trong tp hpcc on thng nh hng quan h bng nhau cng l quan h tngng.

nh ngha 39. Mi lp tng ng sinh ra bi quan h bng nhautrong tp hp cc on thng nh hng c gi l vect.

Vect cha on thng nh hng#

AB c k hiu l [#

AB]. Khikhng quan tm ti cc u mt ca cc on thng nh hng thucvect, thay cho vic k hiu vect bi cc k hiu [

#

AB], [#

CD], [# EF]..., ngi

ta cn k hiu vect bi cc k hiu [#a ], [#

b ], [#c ]...

nh ngha 40. Vect cha cc on thng nh hng-khng c gil vect-khng, k hiu l [#0 ].

nh ngha 41. Hai vect [#a ], [#b ] c gi l cng hng nu mi onthng nh hng thuc vect ny cng hng vi mi on thng nhhng thuc vect kia.

biu th [#a ], [#

b ] cng hng, ta vn s dng cch k hiu saunh ngha 21.

4. Vect, hng v phng ca n 19

nh ngha 42. Hai vect [#a ], [#b ] c gi l ngc hng nu mion thng nh hng thuc vect ny ngc hng vi mi onthng nh hng thuc vect kia.

biu th [#a ], [#

b ] ngc hng, ta vn s dng cch k hiu saunh ngha 22.

nh ngha 43. Hai vect [#a ], [#b ] c gi l cng phng nu chnghoc cng hng hoc ngc hng.

biu th [#a ], [#

b ] cng phng, ta vn s dng cch k hiu saunh ngha 23.

nh ngha 44. di ca vect l di ca cc on thng nhhng thuc vect.

Ch 45. 1) Cc on thng nh hng thuc cng mt vect c di bng nhau.

2) di ca vect [#a ] c k hiu l |[#a ]|.3) Vect c di bng 1 c gi l vect n v.

nh ngha 46. Hai vect [#a ], [#b ] c gi l bng nhau (khc nhau),nu mi on thng nh hng thuc vect ny bng (khc) mi onthng inh hng thuc vect kia.

biu th [#a ], [#

b ] bng nhau (khc nhau) ta vn s dng cch khiu sau nh ngha 24.

nh ngha 47. Hai vect [#a ], [#b ] c gi l i nhau nu mi onthng nh hng thuc vect ny v mi on thng nh hng thucvect kia ngc hng v c di bng nhau.

biu th [#a ], [#

b ] i nhau, ta vit [#a ]=[#b ].

4.2. Cc nh l.

nh l 48. Vi hai on thng nh hng # AB, # CD, ta c

1)#

AB # CD khi v ch khi [# AB] [# CD].2)

#


#


#

AB= # CD khi v ch khi [# AB]= [# CD].

nh l 48 l h qu trc tip ca nh l 35.


nh l 49. Vect [#0 ] cng hng vi mi vect.


nh l 50. Vect [#0 ] ngc hng vi mi vect.


nh l 51. Vect [#0 ] cng phng vi mi vect.


nh l 52. Vi hai im A,B, ta c

1) [#

AB] [# AB].2) [

#

AB] [# BA].nh l 51 l h qu trc tip ca cc nh l 32, 48.

nh l 53. Vi hai vect [# AB], [# CD], ta c [# AB] [# CD] khi v ch khi[

#

AB] [# DC].nh l 53 l h qu trc tip ca cc nh l 33, 48.

nh l 54. Vi ba vect-khc khng [#a ], [#b ], [#c ], ta c

1) Nu [#a ] [#b ]; [#b ] [#c ] th [#a ] [#c ].2) Nu [#a ] [#b ]; [#b ] [#c ] th [#a ] [#c ].3) Nu [#a ] [#b ]; [#b ] [#c ] th [#a ] [#c ].nh l 54 l h qu trc tip ca cc nh l 35, 48.

4.3. Hng v phng ca vect.Theo cc nh l 52, 54, d dng thy rng trong tp hp cc vect-khckhng quan h cng hng l quan h tng ng.

nh ngha 55. Mi lp tng ng sinh ra bi quan h cng hngtrong tp hp cc vect-khc khng c gi l hng ca vect.

Hng ca vect cha vect [#a ] c gi n gin l hng ca vect[#a ].

Thut ng hng ca vect gii thch phn no ngha ca thutng cng hng trong nh ngha 41. C th, hai vect-khc khngcng hng nu chng cng thuc mt hng ca vect.

Nu [#a ] v [#

b ] l hai vect ngc hng th, theo nh l 54, mivect thuc hng ca vect [#a ] ngc hng vi mi vect thuc hngca vect [

#

b ]. Nhn xt ny khng nh s hp l ca nh ngha sau.

5. Hng v phng ca tia 21

nh ngha 56. Hai hng ca vect c gi l ngc nhau nu mivect thuc hng ca vect ny v mi vect thuc hng ca vect kiangc hng.

Theo cc nh l 52, 54, d dng thy rng trong tp hp cc vect-khc khng quan h cng phng cng l quan h tng ng.

nh ngha 57. Mi lp tng ng sinh ra bi quan h cng phngtrong tp hp cc vect-khc khng c gi l phng ca vect.

Phng ca vect cha vect [#a ] c gi n gin l phng cavect [#a ] .

Thut ng phng ca vect gii thch phn no ngha ca thutng cng phng trong nh ngha 43. C th, hai vect-khc khngcng phng nu chng cng thuc mt phng ca vect.

Mi phng ca vect cha hai hng ca vect v l hai hngca vect ngc nhau.

Ch 58. Nu khng c g nhm ln th vect [# AB] c k hiu ngin l

#

AB. Tng t vect [#a ] c k hiu n gin l #a .

5. Hng v phng ca tia

Tip tc s dng cc khi nim: hai on thng nh hng cnghng, hai on thng nh hng ngc hng, d dng i n cckhi nim: hng ca tia, phng ca tia.

5.1. Cc nh ngha.Cho tia Ix v im A thuc Ix. Theo ch 9, A khc I. Do , theo nhl 34, hng ca on thng nh hng

#

IA khng ph thuc vo cchchn im A (h.35).

x

h.35

I A

Ni cch khc, khi A thay i trn tia Ix on thng nh hng#

IAlun thuc mt hng ca on thng nh hng. Nhn xt ny khngnh s hp l ca nh ngha sau.

nh ngha 59. Hng ca on thng nh hng tng thch vi tiaIx l hng ca on thng nh hng

#

IA vi A thuc tia Ix.

nh ngha 60. Hai tia Ix,Jy c gi l cng hng nu cc hngca on thng nh hng tng thch vi chng bng nhau.


biu th Ix,Jy cng hng, ta vn s dng cch k hiu sau nhngha 21.

nh ngha 61. Hai tia Ix,Jy c gi l ngc hng nu cc hngca on thng nh hng tng thch vi chng ngc nhau.

biu th Ix,Jy ngc hng, ta vn s dng cch k hiu nhngha 22.

nh ngha 62. Hai tia Ix,Jy c gi l cng phng nu chng hoccng hng hoc ngc hng.

biu th Ix,Jy cng phng, ta vn s dng cch k hiu saunh ngha 23.

5.2. Cc nh l.

nh l 63. Vi hai tia Ix,Jy, cc mnh sau tng ng (h.36).

1) Ix I y.2) Vi mi im A thuc tia Ix, vi mi im B thuc tia Jy, ta c

#

IA # JB.3) Tn ti im A thuc tia Ix, tn ti im B thuc tia Jy sao cho

#

IA # JB.x

y

h.36

I

J

A

B


nh l 64. Vi hai tia Ix, Jy, cc mnh sau tng ng (h.37).

1) Ix Jy.2) Vi mi im A thuc tia Ix, vi mi im B thuc tia Jy, ta c

#

IA # JB.3) Tn ti im A thuc tia Ix, tn ti im B thuc tia Jy sao cho

#

IA # JB.


x

h.37

y

I A

JB


nh l 65. Vi mi tia Ix, ta c1) Ix Ix.2) Ix Ix, y, Ix l tia i ca tia Ix.

nh l 65 l h qu ca cc nh l 34, 63, 64.

Ch 66. Nu khng c g nhm ln th tia i ca tia Ix c k hiun gin l Ix. ng nhin tia i ca tia Ix l tia Ix (khng chiu l tia Ix).

nh l 67. Vi hai tia Ix,Jy, ta c Ix Jy khi v ch khi Ix Jy.

nh l 67 l h qu trc tip ca cc nh l 34, 35, 63, 64.

nh l 68. Vi ba tia Ix,Jy,Kz, ta c1) Nu Ix Jy;Jy Kz th Ix Kz.2) Nu Ix Jy;Jy Kz th Ix Kz.3) Nu Ix Jy;Jy Kz th Ix Kz.

nh l 68 l h qu trc tip ca cc nh l 35, 63, 64.

nh l 69. Nu hai tia Ix,Jy thuc hai ng thng song song th1) Ix Jy khi v ch khi Ix,Jy cng thuc mt na mt phng b

IJ.2) Ix Jy khi v ch khi Ix,Jy thuc hai na mt phng khc nhau

b IJ.

Chng minh. 1) Ly X Ix;Y Jy. Cc iu kin sau tng ng(h.38).

1) Ix Jy.2) # IX # JY .3) IXY J l hnh thang.


4) X IJY l hnh thang.5) X ,Y / IJ.6) Ix,Jy cng thuc mt na mt phng b IJ.

Ch , theo nh l 63, 1 2; theo nh l 30, 2 3; theo ch 4,3 4; theo b 18, 4 5; hin nhin 5 6.

x

y

x

y

h.38 h.39

I

J

X

Y

I

J

X

Y

2) Ly X Ix;Y Jy. Cc iu kin sau tng ng (h.39).1) Ix Jy.2) # IX # JY .3) # IX # Y J.4) IX JY l hnh thang.5) X IY J l hnh thang.6) X / IJ / I.7) Ix,Jy thuc hai na mt phng khc nhau b IJ.

Ch , theo nh l 64, 1 2; theo nh l 33, 2 3; theo nh l 30,3 4; theo ch 4, 4 5; theo b 18, 5 6; hin nhin 6 7. nh l 70. Cho hai tia Ix,Jy cng hng v khng cng thuc mtng thng; cho l ng thng khng cha I,J v khng song songvi xx, yy. Khi

1) I,J / khi v ch khi hoc cng ct Ix,Jy hoc cng ct Ix,Jy.2) I / / J khi v ch khi hoc cng ct Ix,Jy hoc cng ct

Ix,Jy.

Chng minh. Trn cc tia Ix,Jy theo th t ly cc im A,B. GiX ,Y theo th t l giao im ca v cc ng thng xx, yy.

V Ix Jy nn, theo nh l 63, # IA # JB (1).Theo nh l 34, hoc # IX # IA hoc # IX # IA hoc # JY # JB hoc

# JY # JB (2).


x

y

x

y

x

y

x

y

h.40 h.41

X

Y

X

Y

I

J

A

B

A

B

I

J

1) Cc iu kin sau tng ng (h.40, h.41).

1) I, J / .2) I, J / XY .3) IXY J l hnh thang.4) # IX # JY .

5)

{

# IX # IA# JY # JB{# IX # IA# JY # JB.

6)

{X IxY Jy{X IxY Jy.

7)

{ ct Ix ct Jy{ ct Ix

ct Jy.

Ch , v XY nn 1 2; theo b 18, 2 3; theo nh l 30,3 4; v (1) v (2) nn, theo nh l 35, 4 5; v (1) nn, theo nh l 35,5 4; theo nh l 34, 5 6; v XY nn 6 7.

2) Cc iu kin sau tng ng (h.42, h.43).

1) I / / J.2) I / XY / J.3) IX JY l hnh thang.4) # IX # Y J.


5) # IX # JY .

6)

{

# IX # IA# JY # JB{# IX # IA# JY # JB.

7)

{X IxY Jy{X IxY Jy.

8)

{ ct Ix ct Jy{ ct Ix

ct Jy.

h.42 h.43

xx

yy

x x

yy

X

Y

I

J

A

B

X

Y

A

B

I

J

Ch , v XY nn 1 2; theo b 18, 2 3; theo nh l 30,3 4; theo nh l 33, 4 5; v (1) v (2) nn, theo nh l 35, 5 6;v (1) nn, theo nh l 35, 6 5; theo nh l 34, 6 7; v XY nn7 8. Ch 71. Nu khng c g nhm ln th ng thng cha hai tia inhau Ox,Ox c k hiu n gin l xx.

5.3. Hng v phng ca tia.Theo cc nh l 65, 68, d dng thy rng trong tp hp cc tia quanh cng hng l quan h tng ng.

nh ngha 72. Mi lp tng ng sinh ra bi quan h cng hngtrong tp hp cc tia c gi l hng ca tia .

Hng ca tia cha tia Ix c gi n gin l hng ca tia Ix.Thut ng hng ca tia gii thch ngha ca thut ng cng

6. Hng hn tp, phng hn tp, ng thng nh hng 27

hng trong nh ngha 60. C th, hai tia c gi l cng hng nuchng cng thuc mt hng ca tia.

Nu Ix v Jy l hai tia ngc hng th mi tia thuc hng catia Ix ngc hng vi mi tia thuc hng ca tia Jy. Nhn xt nykhng nh s hp l ca nh ngha sau.

nh ngha 73. Hai hng ca tia c gi l ngc nhau nu mitia thuc hng ca tia ny v mi tia thuc hng ca tia kia ngchng.

Theo cc nh l 65, 68, d dng thy rng trong tp hp cc tiaquan h cng phng cng l quan h tng ng.

nh ngha 74. Mi lp tng ng sinh ra bi quan h cng phngtrong tp hp cc tia c gi l phng ca tia.

Phng ca tia cha tia Ix c gi n gin l phng ca tia Ix.Thut ng phng ca tia gii thch ngha ca thut ng cng

phng trong nh ngha 62. C th, hai tia cng phng nu chngcng thuc mt phng ca tia.

Mi phng ca tia cha hai hng ca tia v l hai hng catia ngc nhau.

6. Hng hn tp, phng hn tp, ng thng nhhng

tin ti khi nim ng thng nh hng ta cn c khi nimphng hn tp.

6.1. Hng hn tp, phng hn tp.

nh ngha 75. on thng nh hng # AB v vect [#a ] c gi lcng hng (ngc hng) nu

#

AB cng hng (ngc hng) vi mion thng nh hng thuc [#a ].

biu th#

AB v [#a ] cng hng (ngc hng), ta vn s dngcch k hiu sau nh ngha 21 (22).

nh ngha 76. on thng nh hng # AB v vect [#a ] c gi lcng phng nu chng hoc cng hng hoc ngc hng.

biu th#

AB v [#a ] cng phng, ta vn s dng cch k hiu saunh ngha 23.


nh ngha 77. on thng nh hng # AB v tia Ox c gi l cnghng (ngc hng) nu

#

AB cng hng (ngc hng) vi mi onthng nh hng thuc hng ca on thng nh hng tng thchvi Ox.

biu th#

AB v Ox cng hng (ngc hng), ta vn s dngcch k hiu sau nh ngha 21 (22).

nh ngha 78. on thng nh hng # AB v tia Ox c gi l cngphng nu chng hoc cng hng hoc ngc hng.

biu th#

AB v Ox cng phng, ta vn s dng cch k hiu saunh ngha 23.

nh ngha 79. Vect [#a ] v tia Ox c gi l cng hng (ngchng) nu mi on thng nh hng thuc [#a ] cng hng (ngchng) vi mi on thng nh hng thuc hng ca on thngnh hng tng thch vi Ox.

biu th [#a ] v Ox cng hng (ngc hng), ta vn s dngcch k hiu sau nh ngha 21 (22).

nh ngha 80. Vect [#a ] v tia Ox c gi l cng phng nu chnghoc cng hng hoc ngc hng.

biu th [#a ] v Ox cng phng, ta vn s dng cch k hiu saunh ngha 23.

D dng thy rng trong tp hp cc on thng nh hng-khckhng, cc vect-khc khng v cc tia quan h cng hng l quan htng ng.

nh ngha 81. Mi lp tng ng sinh bi quan h cng hngtrong tp hp cc on thng nh hng-khc khng, cc vect-khckhng v cc tia c gi l hng hn tp.

Hng hn tp cha on thng nh hng#

AB (cha vect [#a ],cha tia Ox) c gi n gin l hng hn tp

#

AB (hng hn tp[#a ], hng hn tp Ox).

Thut ng hng hn tp gii thch phn no ngha ca thut ngcng hng trong cc nh ngha 75, 77, 79. C th, mt on thngnh hng-khc khng v mt vect-khc khng, mt on thng nhhng-khc khng v mt tia, mt vect-khc khng v mt tia cgi l cng hng nu chng cng thuc mt hng hn tp.

6. Hng hn tp, phng hn tp, ng thng nh hng 29

nh ngha 82. Hai hng hn tp c gi l ngc nhau nu mion thng nh hng-khc khng, mi vect-khc khng v mi tiathuc hng hn tp ny ngc hng vi mi on thng nh hng-khc khng, mi vect-khc khng v mi tia thuc hng hn tp kia.

ng nhin hai on thng nh hng-khc khng cng thucmt hng ca on thng nh hng khi v ch khi chng cng thucmt hng hn tp; hai vect-khc khng cng thuc mt hng cavect khi v ch khi chng cng thuc mt hng hn tp; hai tia cngthuc mt hng ca tia khi v ch khi chng cng thuc mt hnghn tp.

D dng thy rng trong tp hp cc on thng nh hng-khckhng, cc vect-khc khng v cc tia quan h cng phng cng lquan h tng ng.

nh ngha 83. Mi lp tng ng sinh ra bi quan h cng phngtrong tp hp cc on thng nh hng-khc khng, cc vect-khckhng v cc tia c gi l phng hn tp.

Phng hn tp cha on thng nh hng#

AB (cha vect [#a ],cha tia Ox) c gi n gin l phng hn tp

#

AB (phng hn tp[#a ], phng hn tp Ox).

Thut ng phng hn tp gii thch phn no ngha ca thutng cng phng trong cc nh ngh 76, 78, 80. C th mt on thngnh hng-khc khng v mt vect-khc khng, mt on thng nhhng-khc khng v mt tia, mt vect-khc khng v mt tia cgi l cng phng nu chng cng thuc mt phng hn tp.

ng nhin hai on thng nh hng-khc khng cng thucmt phng ca on thng nh hng khi v ch khi chng cngthuc mt phng hn tp; hai vect-khc khng cng thuc mt phngca vect khi v ch khi chng cng thuc mt phng hn tp; hai tiacng thuc mt phng ca tia khi v ch khi chng cng thuc mtphng hn tp.

Mi phng hn tp cha hai hng hn tp v l hai hng hntp ngc nhau.

6.2. ng thng nh hng.Trc ht xin lu rng mt on thng nh hng c gi l thucmt ng thng nu on thng sinh ca n thuc ng thng .


Cho ng thng . Theo nh l 31, cc on thng nh hng-khc khng thuc cng phng. Do chng cng thuc mt phnghn tp. Nhn xt ny khng nh s hp l ca nh ngha sau.

nh ngha 84. Phng hn tp cha cc on thng nh hng-khckhng thuc ng thng c gi l phng hn tp tng thch vi.

D dng thy rng mi on thng nh hng-khc khng thuc hoc cng hng hoc ngc hng vi mi on thng nh hng-khc khng, mi vect-khc khng, mi tia thuc phng hn tp tngthch vi . Nhn xt ny cho ta ba cch nh hng ng thng .

Cch th nht, nh hng bng on thng nh hng-khckhng.

nh ngha 85. Cho ng thng . Ly on thng nh hng-khckhng

#

AB thuc phng hn tp tng thch vi . B hai thnh phn(,

#

AB) c gi l ng thng nh hng bi on thng nhhng

#

AB.

on thng nh hng#

CD thuc c coi l c hng dng (m)khi n cng hng (ngc hng) vi on thng nh hng

#

AB.Theo cc nh l 25, 26, cc on thng nh hng-khng thuc

va cng hng va ngc hng vi on thng nh hng#

AB. Do cc on thng nh hng-khng thuc c coi l va c hngdng va c hng m.

on thng nh hng#

AB c gi l on thng nh hng cang thng nh hng (,

#

AB). Nu khng c g c bit th onthng nh hng ca ng thng nh hng thng c coi l c di bng 1.

Cch th hai, nh hng bng vect-khc khng.

nh ngha 86. Cho ng thng . Ly vect-khc khng [#a ] thucphng hn tp tng thch vi . B hai thnh phn (, [#a ]) c gil ng thng nh hng bi vect [#a ].

on thng nh hng#

CD thuc c coi l c hng dng (m)khi

#

CD cng hng (ngc hng) vi vect [#a ].Theo cc nh l 25, 26, cc on thng nh hng-khng thuc

va cng hng va ngc hng vi vect [#a ]. Do cc on thngnh hng-khng thuc c coi l va c hng dng va c hngm.


Vect [#a ] c gi l vect nh hng ca ng thng nh hng(, [#a ]). Nu khng c g c bit th vect nh hng ca ng thngnh hng thng c coi l vect n v.

Cch th ba, nh hng bng tia.

nh ngha 87. Cho ng thng . Ly tia Ox thuc phng hn tptng thch vi . B hai thnh phn (,Ox) c gi l ng thng nh hng bi tia Ox.

on thng nh hng#

CD thuc c coi l c hng dng (m)khi

#

CD cng hng (ngc hng) vi tia Ox.Theo cc nh l 25, 26, cc on thng nh hng-khng thuc

va cng hng va ngc hng vi tia Ox. Do cc on thng nhhng-khng thuc c coi l va c hng dng va c hng m.

Tia Ox c gi l tia nh hng ca ng thng nh hng(,Ox).

Ch 88. 1) ng thng cha hai tia i nhau Ox,Ox v c nhhng bi tia Ox cng c k hiu n gin l xx.

2) Nu khng c g nhm ln th thut ng on thng nh hngc hng dng (m) c thay bi thut ng on thng nh hngdng (m) v cc thut ng ng thng nh hng bi on thngnh hng

#

AB, ng thng nh hng bi vect [#a ], ng thng nh hng bi tia Ox cng c thay bi thut ng ng thng nhhng .

7. di i s ca on thng nh hng

Trong mc ny, khi nim di i s ca on thng nh hng,cng c quan trng gip ta i s ha hnh hc phng v phng din di on thng c gii thiu.

nh ngha 89. Cho ng thng nh hng . di i s caon thng nh hng

#

CD thuc c k hiu l CD v c xcnh nh sau

CD =

CD nu

#

CD c hng dng CD nu # CD c hng m0 nu

#

CD = #0 .Ch 90. |CD| =CD.nh l 91. Nu hai im C, D cng thuc ng thng nh hng th CD =DC.


Chng minh. C hai trng hp cn xem xt.Trng hp 1. C =D.Theo ch 17, CD = 0=DC.Trng hp 2. C 6=D.Theo ch 90, |CD| = CD = DC = |DC|. Mt khc, theo nh l 32,

#

CD # DC. Do CD v DC khng cng du. Vy CD =DC.

nh l 92. Nu # CD, # EF l hai on thng nh hng-khc khngcng thuc ng thng nh hng th

1)#

CD # EF khi v ch khi CD.EF > 0.2)

#

CD # EF khi v ch khi CD.EF < 0.

Chng minh. Gi # AB l on thng nh hng ca ng thng nhhng .

1) Cc iu kin sau tng ng.

1)#

CD # EF.

2)

{

#

CD # AB# EF # AB{#

CD # AB# EF # AB.

3)

{

CD > 0EF > 0{CD < 0EF < 0.

4) CD.EF > 0.

Ch , theo nh l 35, 1 2; hin nhin, 2 3; hin nhin 3 4.2) Cc iu kin sau tng ng

1)#

CD # EF.

2)

{

#

CD # AB# EF # AB{#

CD # AB# EF # AB.


3)

{

CD > 0EF < 0{CD < 0EF > 0.

4) CD.EF < 0.Ch , theo nh l 35, 1 2; hin nhin, 2 3; hin nhin 3 4.

nh l 93. Nu ba im C,D,M cng thuc ng thng nh hng v i mt khc nhau th

1) M nm ngoi on thng CD khi v ch khi MC.MD > 0.2) M nm trong on thng CD khi v ch khi MC.MD < 0.

Chng minh. 1) Cc iu kin sau tng ng.

1) M nm ngoi on thng CD.2)

#

MC # MD.3) MC.MD > 0.Ch , theo nh l 34, 1 2; theo nh l 92, 2 3.2) Cc iu kin sau tng ng.

1) M nm trong on thng CD.2)

#

MC # MD.3) MC.MD < 0.Ch , theo nh l 34, 1 2; theo nh l 92, 2 3.

nh l 94. Nu ba im A,B,C cng thuc ng thng nh hng th

1) AB= AC+CB (h thc Chasles cho on thng nh hng ).2) AB=CBCA.

Chng minh. 1) C hai trng hp cn xem xt.Trng hp 1. A =B.C hai kh nng xy ra.Kh nng 1.1. A =B=C (h.44).

h.44

A =B=C


Theo ch 17, AB= 0= 0+0= AC+CB.Kh nng 1.2. A =B 6=C (h.45).

h.45

A =B C

Theo ch 17 v nh l 91,

AB= 0= ACAC = ACBC = AC (CB)= AC+CB.Trng hp 2. A 6=B.C nm kh nng xy ra.Kh nng 2.1. C = A.Theo ch 17, AB= 0+AB= AC+CB.Kh nng 2.2. C =B.Theo ch 17, AB= AB+0= AC+CB.Kh nng 2.3. C nm trong on thng AB (h.46).

h.46

A C B

Theo nh l 93, AB.AC > 0 v CA.CB< 0. T , theo nh l 91, suyra AB,AC,CB cng du. Mt khc v C nm trong on thng AB nn,theo nh l 12, AB= AC+CB. Vy AB= AC+CB.

Kh nng 2.4. C thuc tia i ca tia#

AB (h.47).

h.47

AC B

Theo nh l 93, AB.AC < 0 v CA.CB> 0. T , theo nh l 91, suyra AC khc du vi AB,CB. Mt khc v A nm trong on thng BCnn, theo nh l 12, AB=AC+CB. Vy AB= AC+CB.

Kh nng 2.5. C thuc tia i ca tia#

BA (h.48).

h.48

A CB

Theo nh l 93, AB.AC > 0 v CA.CB> 0. T , theo nh l 91, suyra CB khc du vi AB,AC. Mt khc v B nm trong on thng AC


nn, theo nh l 12, AB= ACCB. Vy AB= AC+CB.2) Theo phn 1, AB = AC+CB. T , theo nh l 91, suy ra AB =

CA+CB=CBCA. nh l 95. Nu A,B l hai im phn bit v k l s thc khc 1 th

tn ti duy nht im M thuc ng thng AB sao choMA

MB= k.

Chng minh. C ba trng hp cn xem xt.Trng hp 1. k= 0.

Chn M trng A. Theo ch 17,MA

MB= AAAB

= 0= k.Trng hp 2. k> 0.Chn M ngoi on thng AB sao cho

MAMB

= k. Theo nh l 93,MA.MB> 0.

Vy, theo ch 90,MA

MB= MA

MB

= |MA||MB|

= MAMB

= k.Trng hp 3. k< 0.Chn M trong on thng AB sao cho

MAMB

= k. Theo nh l 93,MA.MB< 0.

Vy, theo ch 90,MA

MB=|MA

MB| = |MA|

|MB|= MA

MB=(k)= k.

NuMA

MB= M

AMB

th, theo nh l 94,

MA

AB= MAMBMA

= MA

MBMA= M

AAB

.

iu c ngha l MA = MA. Do , li theo nh l 94, MA =MM+MA. Vy MM = 0. Ni cch khc, theo ch 17, M =M. Ch 96. Nu ta ni n di i s ca mt on thng nh hngthuc mt ng thng th ng thng c nh hng.

B sau y khng ch gip ta chng minh nh l 98 m cn cvai tr quan trng trong nhiu tnh hung khc.

B 97. (B trung im) Nu im M thuc ng thng AB thcc iu kin sau tng ng.

1) M l trung im ca on thng AB.


2) MA =MB.3) MA =MB.4) 2AM = AB.

Chng minh. (1 2) Hin nhin.(2 3) Nu M khng thuc on thng AB th hoc A thuc on

thng MB hoc B thuc on thng MA. Do , theo nh l 12, MA 6=MB, mu thun. Vy M thuc on thng AB. T , theo nh l 93,suy ra MA.MB< 0. Kt hp vi MA =MB, ta c MA =MB.

(3 4) Theo nh l 91, 2AM = AM+ AM = AMMA. T , ch rng MA =MB, theo nh l 94, suy ra 2AM = AM+MB= AB.

(4 1) Theo nh l 91, AM = 2AM AM = 2AM+MA. T , ch rng 2AM = AB, theo cc nh l 91, 94, suy ra MA = AM = AB+MA =MB. Vy, theo nh l 93 v ch 90, M nm trong on thng AB vMA =MB. Ni cch khc M l trung im ca on thng AB. nh l 98. Nu bn im A,B,C,D cng thuc mt ng thng thcc iu kin sau tng ng.

1) Cc on thng AC,BD c cng trung im.2) AB=DC.3) AD =BC.

Chng minh. (1 2) Gi O l trung im chung ca cc on thngAC,BD. Theo b 97, OA =OC; OB = OD. T , theo nh l 94,suy ra

AB=OBOA =OD+OC =DC.(2 1) Gi O l trung im ca on thng AC. Theo b 97,

OA =OC. T , ch rng AB=DC, theo cc nh l 91, 94, suy raOB=OA+AB=OC+DC = (OCDC)= (OC+CD)=OD

T , ch rng O AC BD, theo b 97, suy ra O l trungim ca on thng BD.

Tm li 1 2.Tng t 1 3.

Chng 2

Hng ca gc

8. Gc gia hai tia

Trong mc ny, mt s kin thc c bn v gc gia hai tia c nhcli.

nh ngha 99. B khng phn bit th t gm hai tia cng gc vkhng cng thuc mt ng thng Ox,Oy c gi l gc gia hai tia,hoc k hiu lxOy hoc k hiu lyOx.nh ngha 100. Giao ca na mt phng b xx cha tia Oy v namt phng b yy cha tia Ox c gi l min trong ca gc gia haitiaxOy.nh ngha 101. B khng phn bit th t gm hai tia trng nhauOx,Oy cng c gi l gc gia hai tia (gc gia hai tia-khng, khicn nhn mnh), k hiu bi mt trong cc cch sau:xOy,yOx,xOx,yOy(h.49).

x= y

h.49

O

Ch 102. Trong hnh 49, k hiu x= y c hiu l tia Ox trng tiaOy.

nh ngha 103. Min trong ca gc gia hai tia-khng l tp hprng.

37

38 2. Hng ca gc

nh ngha 104. B khng phn bit th t gm hai tia i nhauOx,Oy cng vi mt na mt phng b yx cng c gi l gc gia haitia (gc gia hai tia-bt, khi cn nhn mnh), hoc k hiu lxOy hock hiu lyOx (h.50).

y x

h.50

O

nh ngha 105. Na mt phng b yx trong nh ngha 104 l mintrong ca gc gia hai tia-btxOy.Ch 106. 1) ng thng yx trong cc nh ngha 104, 105 l ngthng cha hai tia i nhau Ox,Oy.

2) Khi ta ni ti mt gc gia hai tia-bt, min trong ca n cch nh.

im O c gi l nh, cc tia Ox,Oy c gi l cnh ca gc giahai tiaxOy.

Nu cc im A,B theo th t thuc cc tia Ox,Oy th thay cho cc khiuxOy,yOx ta c th dng cc k hiuAOB,BOA,AOy,yOA,xOB,BOx ch gc gia hai tiaxOy.

Vi s xut hin ca cc khi nim gc gia hai tia-khng, gc giahai tia-bt, thut ng gc gia hai tia mang mt ngha mi, gc giahai tia c th l gc gia hai tia-khc khng v khc bt (hai cnh l haitia khng cng thuc mt ng thng), c th l gc gia hai tia-khng(hai cnh l hai tia trng nhau), c th l gc gia hai tia-khc khng(hai cnh l hai tia khng trng nhau), c th l gc gia hai tia-bt(hai cnh l hai tia i nhau) v cng c th l gc gia hai tia-khc bt(hai cnh l hai tia khng i nhau). V vy, trnh nhm ln, trongcc vn lin quan ti khi nim gc gia hai tia, khi cn thit ta snhn mnh cc thut ng: gc gia hai tia-khc khng v khc bt, gcgia hai tia-khng, gc gia hai tia-khc khng, gc gia hai tia-bt,gc gia hai tia-khc bt.

di phn giao ca min trong ca gc gia hai tiaxOy v ngtrn tm O, bn knh 1 c gi l s o (tnh theo radian) caxOy, khiu l sxOy.

Nu khng c g nhm ln th s o ca gc gia hai tiaxOy ck hiu n gin lxOy. ng nhin 0xOypi.

Ngoi radian ngi ta cn o gc gia hai tia bng ,pi

2radian= 90o

(c l 90 ). Tuy nhin, cho n gin, trong cun sch ny gc gia


hai tia ch c o bng radian. Theo thi quen, cch vit radian cthay bng cch vit n gin hn .

Ch 107. 1)xOy= 0 khi v ch khixOy l gc gia hai tia-khng.2)xOy=pi khi v ch khixOy l gc gia hai tia-bt.3)xOy= pi

2khi v ch khi xx yy.

nh ngha 108. Hai gc gia hai tia xI y, zJt c gi l bng nhaunu s o ca chng bng nhau.

biu th xI y, zJt bng nhau, ta vit xI y= zJt.

nh l 109. Nu tia Oz hoc thuc min trong hoc trng vi mt tronghai cnh ca gc gia hai tiaxOy thxOy=xOz+zOy (h thc Chaslescho gc gia hai tia).

Trong nh l 109, gc gia hai tia xOy c th l gc gia hai tia-khng v cng c th l gc gia hai tia-bt.

Ch 110. Thay cho cch ni tia Oz thuc min trong ca gc gia haitiaxOy ta cn c cc cch ni n gin hn: tia Oz nm trong gc giahai tiaxOy, tia Oz nm gia cc tia Ox,Oy.nh l 111. Cho ng thng xx v hai tia I y,Jz sao cho I,J thucxx. Khi

Cc iu kin sau tng ng (h.51).1) I y Jz.2) Cc tia I y,Jz cng thuc mt na mt phng b xx v

xI y= xJz.3) Cc tia I y,Jz cng thuc mt na mt phng b xx vxI y=xJz.

Cc iu kin sau tng ng (h.52).1) I y Jz.2) Cc tia I y,Jz thuc hai na mt phng khc nhau b xx

v xI y=xJz.3) Cc tia I y,Jz thuc hai na mt phng khc nhau b xx

vxI y= xJz.

40 2. Hng ca gc

y

z

x

xh.51

y

z

x

xh.52

I

J

I

J

Phn 1 ca nh l 111 l h qu trc tip ca phn 1 ca nh l 69v nh l c bn sau: Hai ng thng song song khi v ch khi hai gcng v bng nhau.

Phn 2 ca nh l 111 l h qu trc tip ca phn 2 ca nh l 69v nh l c bn sau: Hai ng thng song song khi v ch khi hai gcso le bng nhau.

nh l 112. Vi hai tia Ox,Oy ta cxOy=xOy.Hai gcxOy,xOy trong nh l 112 c gi l hai gc gia hai tia

i nh. chng minh nh l 114 ta cn c b sau.

B 113. (B tia thuc tia) Nu bn im A,B,C,D cng thuc mtng thng v A 6=C,B 6=D th # AC # BD khi v ch khi hoc # BD # AChoc

#

AC # BD.

Chng minh. iu kin cn. C hai trng hp cn xem xt.Trng hp 1. A =B.C hai kh nng xy ra.Kh nng 1.1. C =D.ng nhin

#

AC # BD.Kh nng 1.2. C 6=D.V

#

AC # BD nn, theo nh l 34, A =B nm ngoi on thng CD.iu c ngha l

#

AC # BD.Trng hp 2. A 6=B.Nu B khng thuc tia

#

AC v A khng thuc tia # BD th A thuc onthng BC v B thuc on thng AD. Do , theo nh l 34,

#

AC # ABv # BD # BA. T , theo cc nh l 33, 35, suy ra # AC # BD, mu thun.Vy c hai kh nng xy ra.

Kh nng 2.1. B thuc tia#

AC.


Ly im M thuc tia # BD. V B thuc tia#

AC v M thuc tia # BDnn A khng thuc on thng BC v B khng thuc on thng DM.Do , theo nh l 93, AB.AC > 0 v BM.BD > 0. V # AC # BD nn,theo nh l 92,

AC

BD> 0. Vy, theo nh l 94, AM.AC = (AB+BM)AC =

AB.AC+BM.AC = AB.AC+BM.BD ACBD

> 0. T , li theo nh l 92,suy ra AM AC. Do , theo nh l 34, A khng thuc on CM. iu c ngha l M thuc tia AC. Vy # BD # AC.

Kh nng 2.2. A thuc tia # BD.Tng t kh nng 2.1,

#

AC # BD.iu kin . Ly im M ng thi thuc cc tia

#

AC,# BD sao cho

min{MA,MB} >max{AC,BD,AB}. Theo nh l 12, A,B,M theo th tnm ngoi cc on thng MC,MD,AB. Do , theo nh l 34,

#

AC #

AM; # BD # BM; # MA # MB. T , theo cc nh l 33, 35, suy ra # AC # BD.

nh l 114. Nu Ix Jz; I y Jt th xI y= zJt.

Chng minh. C ba trng hp cn xem xt.

Trng hp 1.[xI y= 0zJt= 0. Khng mt tnh tng qut gi s xI y= 0.

V xI y= 0 nn, theo ch 107, Ix I y. Do , theo nh l 65, Ix I y. Kt hp vi Ix Jz; I y Jt, theo nh l 68, ta c Jz Jt. T ,theo b 113, suy ra Jz Jt. Li theo ch 107, zJt= 0. Vy xI y= zJt.

Trng hp 2.[xI y=pizJt=pi. Khng mt tnh tng qut gi s xI y=pi.

V xI y=pi nn, theo ch 107, cc Ix, I y i nhau. Do , theo nh l65, Ix I y. Kt hp vi Ix Jz; I y Jt, theo nh l 68, ta c Jz Jt.T , theo nh l 67 v b 113, suy ra Jz Jt. Li theo ch 107,zJt=pi. Vy xI y= zJt.

Trng hp 3.

xI y 6= 0xI y 6=pizJt 6= 0zI t 6=pi.

V Ix Jz; I y Jt nn, theo nh l 31, 63,

[xx zzxx zz[yy ttyy tt.

42 2. Hng ca gc

Do c bn kh nng xy ra.

Kh nng 3.1.{xx zzyy tt. (h.53)

V xI y 6= 0; xI y 6=pi nn I = xxyy. Do I = xxyy= zztt= J. T, ch rng Ix Jz; I y Jt, theo b 113, suy ra Ix Jz; I y Jt.Vy xI y= zJt.

x= z

y= t

y = t

h.53

x = z I = J


t

t

x= zx = z

y

yh.54

J I

V xx zz; yy tt nn I,J cng thuc xx zz; I 6= J. T , ch rngIx Jz, theo b 113, suy ra hoc Jz Ix hoc Ix Jz . Kt hp viI y Jt, theo nh l 111, ta c xI y= zI y= zJt.


y= t

y = t

zz

x x

h.55

I

J


Tng t kh nng 3.2, xI y= zJt.Kh nng 3.4.

{xx zzyy tt.

V xI y 6= 0; xI y 6=pi nn I = xx yy. Do

J nm trong gc xI yJ nm trong gcxI yJ nm trong gcxI yJ nm trong gc xI y.

Khng mt tnh tng qut gi s J nm trong gc xI y (h.56).

y

y

t

t

zz

x

h.56

x

F J

EI

V xx, yy ct nhau v tt yy; zz xx nn xx, tt ct nhau v yy, zzct nhau. t E = xx tt;F = yy zz.

V J nm trong gc xI y nn J thuc na mt phng b yy chatia Ix. T , ch rng J thuc tt v tt yy, suy ra E thuc tia Ix.iu c ngha l Ex Ix. Do , theo b 113, Ix Ex. Kt hp viIx Jz, theo nh l 68, ta c Ex Jz. Vy, theo nh l 111, xEt= zJt.

Tng t nh trn F thuc tia I y v Fy Jt. V zz xx nn F,J / xx.V F thuc tia I y nn I thuc tia Fy. T , theo nh l 70, suy ra Ethuc tia Jt. Do J thuc tia Et. iu c ngha l JtEt. Vy, theob 113, Et Jt. Kt hp vi I y Jt, theo nh l 68, ta c Et I y.T , theo nh l 111, suy ra xEt= xI y.

Tm li xI y= zJt. Hai gc xI y, zJt trong nh l 114 c gi l hai gc gia hai tia c

cc cnh tng ng cng hng.

nh l 115. Nu Ix Jz; I y Jt th xI y= zJt.

Chng minh. Theo nh l 112, xI y=xI y.Theo nh l 65, Ix Ix; I y I y. T , ch rng Ix Jz; I y

Jt, theo nh l 68, suy ra Ix Jz; I y Jt.

44 2. Hng ca gc

Do , theo nh l 114, xI y = zJt.Vy xI y= zJt. Hai gc xI y, zJt trong nh l 115 c gi l hai gc gia hai tia c

cc cnh tng ng ngc hng.

9. Gc nh hng gia hai tia v cc vn c lin quan

Trong mc 8, khi nh ngha gc gia hai tia ta khng phn bit th thai cnh ca n. Trong mc ny, ta lm quen vi mt khi nim mi:gc gia hai tia m th t hai cnh ca n phn bit.

9.1. Gc nh hng gia hai tia.

nh ngha 116. B c phn bit th t gm hai tia cng gc v khngi nhau (Ox,Oy) c gi l gc nh hng gia hai tia-khc bt, cngk hiu l (Ox,Oy).

Khi cc tia Ox,Oy trng nhau, gc nh hng gia hai tia-khc bt(Ox,Oy) c gi l gc nh hng gia hai tia-khng, cn k hiu bimt trong cc cch sau: (Oy,Ox), (Ox,Ox), (Ox,Oy).

nh ngha 117. Min trong ca gc nh hng gia hai tia-khc bt(Ox,Oy) l min trong ca gc gia hai tiaxOy.nh ngha 118. B c phn bit th t gm hai tia cng gc v inhau (Ox,Oy) cng vi mt na mt phng b yx c gi l gc nhhng gia hai tia-bt, cng k hiu l (Ox,Oy).

nh ngha 119. Na mt phng b yx trong nh ngha 119 c gil min trong ca gc nh hng gia hai tia-bt (Ox,Oy).

Ch 120. Khi ta ni ti mt gc nh hng gia hai tia-bt, mintrong ca n c ch nh.

im O c gi l nh, cc tia Ox, Oy theo th t c gi l cnhu, cnh cui ca gc nh hng gia hai tia (Ox,Oy).

Nu cc im A,B theo th t thuc cc tia Ox,Oy th thay cho khiu (Ox,Oy) ta c th dng cc k hiu (

#

OA,#

OB), (#

OA,Oy), (Ox,#

OB) ch gc nh hng gia hai tia (Ox,Oy).

Cng vi thut ng gc nh hng gia hai tia-khng, ta c thutng gc nh hng gia hai tia-khc khng. Cng vi thut ng gcnh hng gia hai tia-khng v thut ng gc nh hng gia haitia-bt, ta c thut ng gc nh hng gia hai tia-khc khng v khc

9. Gc nh hng gia hai tia v cc vn c lin quan 45

bt. Khi khng cn phn bit, cc thut ng gc nh hng gia haitia-khng, gc nh hng gia hai tia-khc khng, gc nh hng giahai tia-bt, gc nh hng gia hai tia-khc bt, gc nh hng giahai tia-khc khng v khc bt cng c thay bng thut ng gc nhhng gia hai tia.

9.2. C s, tia c s ca gc nh hng gia hai tia-khc bt ccng nh.

nh ngha 121. Vi hai s thc dng khc nhau a,b, tp hp B(Ox,Oy)= {(Ox,a); (Oy,b); (Ox,a); (Oy,b)} c gi l c s ca gc nh hnggia hai tia khc bt (Ox,Oy).

nh ngha 122. Cc phn t ca B(Ox,Oy) c gi l cc tia c sca gc nh hng gia hai tia-khc bt (Ox,Oy).

Mi tia c s ca gc nh hng gia hai tia-khc bt (Ox,Oy) chai thnh phn, thnh phn th nht-phn hnh, thnh phn th hai-phn s.

Cho hai gc nh hng gia hai tia-khc bt (Ox,Oy), (Oz,Ot) vicc c s tng ng l

B(Ox,Oy)= {(Ox,a); (Oy,b); (Ox,a); (Oy,b)};B(Oz,Ot)= {(Oz, c); (Ot,d); (Oz,c); (Ot,d)}.

Nu a,b, c,d i mt khc nhau th tp hp B(Ox,Oy)B(Oz,Ot)=;.Nhn xt ny khng nh s hp l ca nh ngha sau.

nh ngha 123. Vi B(Ox,Oy)= {(Ox,a); (Oy,b); (Ox,a); (Oy,b)},B(Oz,Ot) = {(Oz, c); (Ot,d); (Oz,c); (Ot,d)} v a,b, c,d l bn s thcdng i mt khc nhau, tp hp B(Ox,Oy,Oz,Ot)=B(Ox,Oy){B(Oz,Ot)c gi l c s ca hai gc nh hng gia hai tia-khc bt (Ox,Oy), (Oz,Ot).

nh ngha 124. Cc phn t ca B(Ox,Oy,Oz,Ot) cng c gi l cctia c s ca hai gc nh hng gia hai tia-khc bt (Ox,Oy), (Oz,Ot).

ng nhin mi tia c s ca hai gc nh hng gia hai tia-khcbt (Ox,Oy), (Oz,Ot) cng c hai thnh phn, thnh phn th nht-phnhnh, thnh phn th hai-phn s.

nh ngha 125. Cho hai gc nh hng gia hai tia-khc bt (Ox,Oy),(Oz,Ot) vi c s l B(Ox,Oy,Oz,Ot). Lymt phn t bt k ca B(Ox,Oy,Oz,Ot) v k hiu phn hnh ca n l Om. C hai na mt phng bmm. Hnh gm mt trong hai na mt phng b mm v tia Om c

46 2. Hng ca gc

gi l na mt phng b mm m rng bi tia Om. C hai na mtphng mm m rng bi tia Om. K hiu bi (M) mt trong hai namt phng b mm m rng bi tia Om. Tn ti ng ba phn t caB(Ox,Oy,Oz,Ot) khc phn t c phn hnh l Om v phn hnh cachng thuc (M). K hiu phn hnh ca ba phn t l On,Op,Oqtheo quy tc sau:mOnmOpmOq (h.57, h.58, h.59, h.60, h.61).

p

p

q

q

mm

n

n

h.57

m = n m= n

q

q

p

ph.58

m = n m= n

p= q

p = q

m= n= pm = n = p

q

q

h.59 h.60

m = n = p = q m= n= p= q

h.61

O O

O O

O

Cch k hiu phn hnh ca cc phn t ca B(Ox,Oy,Oz,Ot) nhtrn c gi l cch k hiu chun ca B(Ox,Oy,Oz,Ot).

V c tm cch chn tia Om v vi mi tia Om li c hai na mtphng b mm nn c mi su cch k hiu chun ca B(Ox,Oy,Oz,Ot).

Ch 126. V c rt nhiu b bn s thc dng i mt khc nhaunn hai gc nh hng gia hai tia-khc bt (Ox,Oy), (Oz,Ot) c rtnhiu c s. Tuy nhin phn hnh ca cc tia c s ca hai c s bt ktng ng trng nhau. Do nu khng k ti s sai khc v mt khiu th ch c mi su cch k hiu chun ca B(Ox,Oy,Oz,Ot).

9.3. S khng trng lp, s trng lp ca hai gc nh hnggia hai tia-khc bt c cng nh.


nh ngha 127. Hai gc nh hng gia hai tia-khc bt (Ox,Oy),(Oz,Ot) c gi l khng trng lp nu cc ng thng xx, yy, zz, tti mt khc nhau (h.57).

nh ngha 128. Hai gc nh hng gia hai tia-khc bt (Ox,Oy),(Oz,Ot) c gi l trng lp nu cc ng thng xx, yy, zz, tt khngi mt khc nhau.

Khi hai gc nh hng gia hai tia-khc bt (Ox,Oy), (Oz,Ot) trnglp, c 4 kh nng xy ra (h.58, h.59, h.60, h.61).

Trng lp loi mt: Trong cc ng thng xx, yy, zz, tt c ngmt b hai ng thng trng nhau (h.58).

Trng lp loi hai: Trong cc ng thng xx, yy, zz, tt c nghai b hai ng thng trng nhau (h.59).

Trng lp loi ba: Trong cc ng thng xx, yy, zz, tt c ngmt b ba ng thng trng nhau (h.60).

Trng lp loi bn: Cc ng thng xx, yy, zz, tt trng nhau(h.61).

Ch 129. Nu (Ox,Oy), (Oz,Ot) khng trng lp th (Ox,Oy), (Oz,Ot)l gc nh hng gia hai tia-khc khng. Nu hoc (Ox,Oy) hoc(Oz,Ot) l gc nh hng gia hai tia-khng th (Ox,Oy), (Oz,Ot) trnglp.

9.4. Ngun v ct tuyn ca hai gc nh hng gia hai tia-khc bt c cng nh.

nh ngha 130. Cho hai gc nh hng gia hai tia-khc bt (Ox,Oy),(Oz,Ot) vi c s l B(Ox,Oy,Oz,Ot). K hiu chun ca B(Ox,Oy,Oz,Ot)nh cch k hiu trong nh ngha 125. K hiu N1,N2,N3,N4,N5,N6,N7,N8l cc tp con bn phn t ca B(Ox,Oy,Oz,Ot) sao cho bn phn tthuc cc tp con ny theo th t c phn hnh l Om,On,Op,Oq;On,Op, Oq,Om; Op,Oq,Om,On; Oq,Om,On,Op;Om,On,Op,Oq;On,Op,Oq,Om;Op,Oq,Om,On;Oq,Om,On,Op. Mi tpcon Ni (i = 1,2,3, . . . ,8) c gi l ngun ca (Ox,Oy), (Oz,Ot) vi c sl B(Ox,Oy,Oz,Ot).

Ch 131. 1) Khi ta ni ti ngun ca hai gc nh hng gia haitia-khc bt (Ox,Oy), (Oz,Ot), B(Ox,Oy,Oz,Ot) c ch nh.

2) Nu khng k n s sai khc v k hiu, cc ngun ca (Ox,Oy),(Oz,Ot) c xc nh mt cch duy nht khng ph thuc vo cch khiu chun ca B(Ox,Oy,Oz,Ot).

48 2. Hng ca gc

3) Cch k hiu trong nh ngha 130 c gi l cch k hiu ccngun ca (Ox,Oy), (Oz,Ot) tng thch vi cch k hiu chun ca B(Ox,Oy,Oz,Ot) trong nh ngha 125.

4) Mi ngun ca (Ox,Oy), (Oz,Ot) c hai phn t thuc B(Ox,Oy) vhai phn t thuc B(Oz,Ot).

nh ngha 132. Mt ngun ca hai gc nh hng gia hai tia-khcbt (Ox,Oy), (Oz,Ot) c gi l thc nu n khng cha hai phn t cphn hnh l hai tia i nhau.

D dng thy rng mt ngun ca (Ox,Oy), (Oz,Ot) l thc khi vch khi tn ti nhng ng thng khng i qua O v ct tt c ccphn hnh ca cc phn t trong ngun.

nh ngha 133. Mt ngun ca hai gc nh hng gia hai tia-khcbt (Ox,Oy), (Oz,Ot) c gi l o nu n cha hai phn t c phnhnh l hai tia i nhau.

D dng thy rng mt ngun ca (Ox,Oy), (Oz,Ot) l o khi v chkhi khng tn ti nhng ng thng khng i qua O v ct tt c ccphn hnh ca cc phn t trong ngun.

nh ngha 134. Tch ca tt c cc phn s ca cc phn t thuc mtngun ca hai gc nh hng gia hai tia-khc bt (Ox,Oy), (Oz,Ot)c gi l c s ca ngun .

nh ngha 135. Du ca c s ca mt ngun ca hai gc nhhng gia hai tia khc-bt (Ox,Oy), (Oz,Ot) c gi l du ca ngun.

nh ngha 136. Mt ngun ca hai gc nh hng gia hai tia-khcbt (Ox,Oy), (Oz,Ot) c gi l dng (m) nu du ca n dng (m).

nh ngha 137. Hai ngun ca hai gc nh hng gia hai tia-khcbt (Ox,Oy), (Oz,Ot) c gi l cng du nu du ca chng hoc cngdng hoc cng m.

nh ngha 138. Hai ngun ca hai gc nh hng gia hai tia-khcbt (Ox,Oy), (Oz,Ot) c gi l tri du nu du ca chng mt dngv mt m.

nh ngha 139. Hai ngun ca hai gc nh hng gia hai tia-khcbt (Ox,Oy), (Oz,Ot) c gi l i nhau nu chng khng c phn tchung.


Ch 140. S dng cch k hiu trong cc nh ngha 125, 130, tathy, vi i = 1,2,3, ...,8, Ni, Ni+4 (N9 =N1;N10 =N2;N11 =N3;N12 =N4) lhai ngun i nhau, phn hnh ca cc phn t ca ngun ny tngng l tia i ca phn hnh ca cc phn t ca ngun kia. Vy haingun i nhau l hai ngun cng du .

nh ngha 141. Hai ngun ca hai gc nh hng gia hai tia-khcbt (Ox,Oy), (Oz,Ot) c gi l k nhau nu chng c chung ng baphn t.

Ch 142. S dng cch k hiu trong cc nh ngha 125, 130, tathy, vi i = 1,2,3, . . . ,8, Ni, Ni+1 (N9 = N1) l hai ngun k nhau, mimt trong hai tp hp Ni \ (Ni Ni+1) v Ni+1 \ (Ni Ni+1) ch c ngmt phn t v phn hnh ca phn t trong tp hp ny l tia i caphn hnh ca phn t trong tp hp kia, ta gi chng l hai tia inhau ca hai ngun k nhau. Vy hai ngun k nhau l hai ngun tridu.

nh ngha 143. Hai ngun ca hai gc nh hng gia hai tia-khcbt (Ox,Oy), (Oz,Ot) c gi l cch nhau nu chng c chung ng haiphn t.

Ch 144. S dng cch k hiu trong cc nh ngha 125, 130, tathy, vi i = 1,2,3, . . . ,8, Ni, Ni+2 (N9 = N1;N10 = N2) l hai ngun cchnhau, mi mt trong hai tp hp Ni \ (Ni Ni+2) v Ni+2 \ (Ni Ni+2)c ng hai phn t v phn hnh ca hai phn t trong tp hp nytng ng l tia i ca phn hnh ca hai phn t trong tp hp kia,ta gi chng l hai b hai tia i nhau ca hai ngun cch nhau. Vyhai ngun cch nhau l hai ngun cng du.

nh ngha 145. ng thng khng i qua O v ct cc ngthng xx, yy c gi l ct tuyn ca gc nh hng gia hai tia-khc bt (Ox,Oy).

Ch 146. V ch c mt giao im vi ng thng xx v mt giaoim vi ng thng yy nn tn ti ng hai phn t ca B(Ox,Oy)m phn hnh ca chng b ct bi .

nh ngha 147. Nu l ct tuyn ca gc nh hng gia hai tia-khc bt (Ox,Oy) v xx = X ; yy = Y th on thng nh hng# XY c gi l vt ca gc nh hng gia hai tia (Ox,Oy) trn .

Vt ca gc nh hng gia hai tia-khc bt (Ox,Oy) trn ct tuyn ca (Ox,Oy) v di i s ca n theo th t c k hiu l#(Ox,Oy) v (Ox,Oy).

50 2. Hng ca gc

nh ngha 148. c s ca ct tuyn ca gc nh hng gia haitia-khc bt (Ox,Oy) vi c s B(Ox,Oy)= {(Ox,a); (Oy,b); (Ox,a); (Oy,b)},c k hiu l [Ox,Oy] v c xc nh nh sau

[Ox,Oy]=

a.b khi X Ox,Y Oya.(b) khi X Ox,Y Oy(a).b khi X Ox,Y Oy(a).(b) khi X Ox,Y Oy.

Ch 149. Khi ta ni ti c s ca ct tuyn ca gc nh hng giahai tia-khc bt (Ox,Oy), B(Ox,Oy) c ch nh.

nh ngha 150. ng thng khng i qua O v ct cc ngthng xx, yy, zz, tt c gi l ct tuyn ca hai gc nh hng giahai tia-khc bt (Ox,Oy), (Oz,Ot).

ng nhin l ct tuyn ca hai gc nh hng gia hai tia-khc bt (Ox,Oy), (Oz,Ot) khi v ch khi va l ct tuyn ca (Ox,Oy)va l ct tuyn ca (Oz,Ot).

Nu l ct tuyn ca hai gc nh hng gia hai tia-khc bt(Ox,Oy), (Oz,Ot) th, theo ch 146, c bn phn t ca B(Ox,Oy,Oz,Ot),hai phn t thuc B(Ox,Oy) v hai phn t thuc B(Oz,Ot), m phnhnh ca chng b ct bi . Do bn phn t ny cng thuc mtngun thc. Nhn xt ny khng nh s hp l ca nh ngha sau.

nh ngha 151. Ngun thc ca hai gc nh hng gia hai tia-khcbt (Ox,Oy), (Oz,Ot) m phn hnh ca cc phn t ca n cng b ctbi ct tuyn ca (Ox,Oy), (Oz,Ot) c gi l ngun sinh ca .

nh ngha 152. Du ca ngun sinh ca ct tuyn ca hai gc nhhng gia hai tia-khc bt (Ox,Oy), (Oz,Ot) c gi l du ca .

nh ngha 153. Tch [Ox,Oy].[Oz,Ot] c gi l c s ca cttuyn ca hai gc nh hng gia hai tia-khc bt (Ox,Oy), (Oz,Ot).

Ch 154. [Ox,Oy].[Oz,Ot] chnh l c s ca ngun sinh ca. Do du ca i vi hai gc nh hng gia hai tia-khc bt(Ox,Oy), (Oz,Ot) chnh l du ca [Ox,Oy].[Oz,Ot].

nh ngha 155. Ct tuyn ca hai gc nh hng gia hai tia-khcbt (Ox,Oy), (Oz,Ot) c gi l dng (m) nu du ca n i vi haigc nh hng gia hai tia-khc bt (Ox,Oy), (Oz,Ot) l dng (m).

nh ngha 156. Hai ct tuyn , ca hai gc nh hng gia haitia-khc bt (Ox,Oy), (Oz,Ot) c gi l cng du nu du ca chng


i vi hai gc nh hng gia hai tia-khc bt (Ox,Oy), (Oz,Ot) hoccng dng hoc cng m.

nh ngha 157. Hai ct tuyn , ca hai gc nh hng gia haitia-khc bt (Ox,Oy), (Oz,Ot) c gi l tri du nu du ca chng ivi hai gc nh hng gia hai tia-khc bt (Ox,Oy), (Oz,Ot)mt dngv mt m.

nh ngha 158. Hai ct tuyn , ca hai gc nh hng gia haitia-khc bt (Ox,Oy), (Oz,Ot) c gi l cng ngun nu chng c cngmt ngun sinh.

ng nhin hai ct tuyn cng ngun l hai ct tuyn cng du.

nh ngha 159. Hai ct tuyn , ca hai gc nh hng gia haitia-khc bt (Ox,Oy), (Oz,Ot) c gi l i ngun nu cc ngun sinhca chng l hai ngun i nhau.

Theo ch 140, hai ct tuyn i ngun l hai ct tuyn cng du.

nh ngha 160. Hai ct tuyn , ca hai gc nh hng gia haitia-khc bt (Ox,Oy), (Oz,Ot) c gi l k ngun nu cc ngun sinhca chng l hai ngun k nhau.

Theo ch 142, hai ct tuyn k ngun l hai ct tuyn tri du.

nh ngha 161. Hai ct tuyn , ca hai gc nh hng gia haitia-khc bt (Ox,Oy), (Oz,Ot) c gi l cch ngun nu cc ngun sinhca chng l hai ngun cch nhau.

Theo ch 144, hai ct tuyn cch ngun l hai ct tuyn cng du.

9.5. Cc nh l v ct tuyn ca hai gc nh hng gia haitia-khc bt cng nh.Trc khi pht biu v chng minh cc nh l c bn v ct tuyn cahai gc nh hng gia hai tia-khc bt c cng nh xin gii thiumt b quan trng.

B 162. (B bn im) Nu bn im A,B,C,D cng thuc mt tiagc A m rng v AB AC AD th # AB # AC # AD # BC # BD # CD.Chng minh. C tm trng hp cn xem xt: 1) A 6= B 6= C 6= D, 2)A = B 6= C 6= D, 3) A 6= B = C 6= D, 4) A 6= B 6= C = D, 5) A 6= B = C = D, 6)A =B 6=C =D, 7) A =B=C 6=D, 8) A =B=C =D.

cho n gin ta ch chng minh b trong trng hp 1, trng

52 2. Hng ca gc

hp phc tp nht (php chng minh b trong cc trng hp cn litng t v n gin hn php chng minh b trong trng hp 1)(h.62).

h.62

A DB C

V AB AC AD nn, theo nh l 34,#

DC # DB; # CB # CA (1);#

AB # AC; # AB # AD; # BC # BD (2).

T (1), theo cc nh l 33, 35, suy ra#

CD # BD; # BC # AC (3).T (2) v (3), theo nh l 35, suy ra

#

AB # AC # AD # BC # BD #

CD.

nh l 163. Nu , l hai ct tuyn cng ngun ca hai gc nhhng gia hai tia-khc bt (Ox,Oy), (Oz,Ot) th

(Ox,Oy).(Oz,Ot).(Ox,Oy).(Oz,Ot) 0.

Chng minh. V , l hai ct tuyn cng ngun ca (Ox,Oy), (Oz,Ot)nn chng c chung ngun sinh. K hiu phn hnh ca cc phn tthuc ngun sinh chung ca , l Om,On,Op,Oq sao cho mOn mOpmOq (h.63).

Gi s theo th t ct Om,On,Op,Oq ti M,N,P,Q; theo tht ct Om,On,Op,Oq ti M,N ,P ,Q.

VmOnmOpmOq nn MN MP MQ;MN MP MQ.Vy, theo b 162,{

# MN # MP # MQ # NP # NQ # PQ#

MN # MP # MQ # N P # N Q # P Q. (1)

C hai mi t trng hp cn xem xt. Tuy nhin v XY .ZT =XY .TZ =Y X .ZT =Y X .TZ vi mi b bn im thng hng X ,Y ,Z,Tv v vai tr ca (Ox,Oy), (Oz,Ot) bnh ng trong php chng minh nynn ta ch cn xem xt ba trng hp.

Trng hp 1. #(Ox,Oy)= # MN; #(Oz,Ot)= # PQ.


h.63

m

n

p

q

O

M

N

P

Q

M

N

P

Q

ng nhin#

(Ox,Oy)= # MN ; #(Oz,Ot)= # P Q.Kt hp cc ng thc trn v (1), theo cc nh l 35, 92, suy ra

(Ox,Oy).(Oz,Ot)=MN.PQ 0;(Ox,Oy).(Oz,Ot)=MN .P Q 0.

Do (Ox,Oy).(Oz,Ot).(Ox,Oy).(Oz,Ot) 0.Trng hp 2. #(Ox,Oy)= # MP; #(Oz,Ot)= # NQ.ng nhin

#

(Ox,Oy)= # MP ; #(Oz,Ot)= # N Q.Kt hp cc ng thc trn v (1), theo cc nh l 35, 92, suy ra

(Ox,Oy).(Oz,Ot)=MP.NQ 0;(Ox,Oy).(Oz,Ot)=MP .N Q 0.

Do (Ox,Oy).(Oz,Ot).(Ox,Oy).(Oz,Ot) 0.Trng hp 3.#(Ox,Oy)= # MQ; #(Oz,Ot)= # NP.ng nhin

#

(Ox,Oy)= # MQ; #(Oz,Ot)= # N P .Kt hp cc ng thc trn v (1), theo cc nh l 35, 92, suy ra

(Ox,Oy).(Oz,Ot)=MQ.NP 0;(Ox,Oy).(Oz,Ot)=MQ.N P 0.

Do (Ox,Oy).(Oz,Ot).(Ox,Oy).(Oz,Ot) 0.Tm li, trong mi trng hp ta u c


54 2. Hng ca gc

nh l 164. Nu , l hai ct tuyn i ngun ca hai gc nhhng gia hai tia-khc bt (Ox,Oy), (Oz,Ot) th


Chng minh. V , l hai ct tuyn i ngun ca (Ox,Oy), (Oz,Ot)nn ngun sinh ca chng i nhau. K hiu phn hnh ca cc phnt thuc ngun sinh ca l Om,On,Op,Oq sao cho mOn mOp mOq. D thy phn hnh ca cc phn t thuc ngun sinh ca lOm,On,Op,Oq (h.64).

m

n p

p

q

q

n

m

h.64

O

N

Q

P

MM

N

P

Q


VmOnmOpmOq nn MN MP MQ.Cng vmOn mOp mOq nn, theo nh l 112, mOn mOp mOq. Do MN MP MQ.Vy, theo b 162,{


MN # MP # MQ # N P # N Q # P Q.T , tng t php chng minh nh l 163, suy ra



nh l 165. Nu , l hai ct tuyn k ngun ca hai gc nh hnggia hai tia-khc bt (Ox,Oy), (Oz,Ot) th


Chng minh. V , l hai ct tuyn k ngun ca (Ox,Oy), (Oz,Ot)nn ngun sinh ca chng k nhau. K hiu phn hnh ca cc phnt thuc ngun sinh ca l Om,On,Op, Oq sao cho mOn mOp mOq v Om,Om l hai tia i nhau ca hai ngun sinh k nhau ca,. D thy phn hnh ca cc phn t thuc ngun sinh ca lOm,On,Op,Oq (h.65).

mm

np

q

h.65

Q

N

M

O

P

M

Q P

N


VmOnmOpmOq nn MN MP MQ.Cng vmOn mOp mOq nn pimOn pimOp pimOq. Do

, theo nh l 109, mOq mOp mOn. iu c ngha l MQ MP MN .

Vy, theo b 162,{# MN # MP # MQ # NP # NQ # PQ#

MQ # MP # MN # QP # QN # P N . (1)

Cng nh php chngminh nh l 163, ta ch cn xem xt ba trnghp.

56 2. Hng ca gc

Trng hp 1. #(Ox,Oy)= # MN; #(Oz,Ot)= # PQ.ng nhin

#

(Ox,Oy)= # MN ; #(Oz,Ot)= # P Q.Kt hp cc ng thc trn v (1), theo cc nh l 33, 35, 92, suy ra

(Ox,Oy).(Oz,Ot)=MN.PQ 0;

(Ox,Oy).(Oz,Ot)=MN .P Q 0.Do (Ox,Oy).(Oz,Ot).(Ox,Oy).(Oz,Ot) 0.Trng hp 2. #(Ox,Oy)= # MP; #(Oz,Ot)= # NQ.ng nhin

#

(Ox,Oy)= # MP ; #(Oz,Ot)= # N Q.Kt hp cc ng thc trn v (1), theo cc nh l 33, 35, 92, suy ra

(Ox,Oy).(Oz,Ot)=MP.NQ 0;

(Ox,Oy).(Oz,Ot)=MP .N Q 0.Do (Ox,Oy).(Oz,Ot).(Ox,Oy).(Oz,Ot) 0.Trng hp 3.#(Ox,Oy)= # MQ; #(Oz,Ot)= # NP.ng nhin

#

(Ox,Oy)= # MQ; #(Oz,Ot)= # N P .Kt hp cc ng thc trn v (1), theo nh l 33, 35, 92, suy ra

(Ox,Oy).(Oz,Ot)=MQ.NP 0;

(Ox,Oy).(Oz,Ot)=MQ.N P 0.Do (Ox,Oy).(Oz,Ot).(Ox,Oy).(Oz,Ot) 0.Tm li, trong mi trng hp ta u c


nh l 166. Nu , l hai ct tuyn cch ngun ca hai gc nhhng gia hai tia-khc bt (Ox,Oy), (Oz,Ot) th


Chng minh. V , l hai ct tuyn cch ngun ca (Ox,Oy), (Oz,Ot)nn ngun sinh ca chng cch nhau. K hiu phn hnh ca cc phnt thuc ngun sinh ca l Om,On,Op,Oq sao chomOnmOpmOqv Om,Om;On,On l hai b hai tia i nhau ca hai ngun sinh cchnhau ca ,. D thy phn hnh ca cc phn t thuc ngun sinhca l Om,On,Op,Oq (h.66).


n

n

m

p

q

q

m

p

h.66

O M

P

Q

N

P

Q

N

M


VmOnmOpmOq nn MN MP MQ.Mt khc vmOnmOpmOqpi nnmOpmOnmOqmOn

pimOn.T , theo nh l 109, suy ra nOpnOqnOm. iu c ngha

l pinOp pinOq pinOm. Li theo nh l 109, nOm nOq nOp. Do N M N Q N P .Vy, theo b 162,{


N M # N Q # N P # MQ # MP # QP . (1)

Cng nh php chng minh nh l 163, ta cng ch cn xem xt batrng hp.

Trng hp 1. #(Ox,Oy)= # MN; #(Oz,Ot)= # PQ.ng nhin

#

(Ox,Oy)= # MN ; #(Oz,Ot)= # P Q.Kt hp cc ng thc trn v (1), theo cc nh l 33, 35, 92, suy ra

(Ox,Oy).(Oz,Ot)=MN.PQ 0;(Ox,Oy).(Oz,Ot)=MN .P Q 0.

58 2. Hng ca gc

Do (Ox,Oy).(Oz,Ot).(Ox,Oy).(Oz,Ot) 0.Trng hp 2. #(Ox,Oy)= # MP; #(Oz,Ot)= # NQ.ng nhin

#

(Ox,Oy)= # MP ; #(Oz,Ot)= # N Q.Kt hp cc ng thc trn v (1), theo cc nh l 35, 92, suy ra

(Ox,Oy).(Oz,Ot)=MP.NQ 0;(Ox,Oy).(Oz,Ot)=MP .N Q 0.

Do (Ox,Oy).(Oz,Ot).(Ox,Oy).(Oz,Ot) 0.Trng hp 3.#(Ox,Oy)= # MQ; #(Oz,Ot)= # NP.ng nhin #(Ox,Oy)= # MQ; #(Oz,Ot)= # N P .Kt hp cc ng thc trn v (1), theo cc nh l 35, 92, suy ra

(Ox,Oy).(Oz,Ot)=MQ.NP 0;(Ox,Oy).(Oz,Ot)=MQ.N P 0.

Do (Ox,Oy).(Oz,Ot).(Ox,Oy).(Oz,Ot) 0.Tm li, trong mi trng hp ta u c


nh l 167. Nu , l cc ct tuyn ca hai gc nh hng gia haitia-khc bt (Ox,Oy), (Oz,Ot) th

1) (Ox,Oy).(Oz,Ot).(Ox,Oy).(Oz,Ot) 0 khi , cng du.2) (Ox,Oy).(Oz,Ot).(Ox,Oy).(Oz,Ot) 0 khi , tri du.

Chng minh. K hiu phn hnh ca cc phn t ca B(Ox,Oy,Oz,Ot)l Om,On,Op,Oq,Om,On,Op,Oq, nh cch k hiu trong nh ngha125. K hiu cc ngun ca B(Ox,Oy,Oz,Ot) l N1, ,N8, nh cch khiu trong nh ngha 130.

Tu theo tng trng hp s cc ngun thc v s cc ngun o caB(Ox,Oy,Oz,Ot) thay i nh sau

p

p

q

q

mm

n

n

h.67

m = n m= n

q

q

p

ph.68

O O


m = n m= n

p= q

p = q

m= n= pm = n = p

q

q

h.69 h.70

m = n = p = q m= n= p= q

h.71

O O

O

(Ox,Oy), (Oz,Ot) khng trng lp. Tm ngun thc: N1,N2,N3,N4, N5,N6,N7,N8; khng ngun o (h.67).

(Ox,Oy), (Oz,Ot) trng lp loi mt. Su ngun thc : N1,N3,N4, N5,N7,N8; hai ngun o: N2,N6 (h.68).

(Ox,Oy), (Oz,Ot) trng lp loi hai. Bn ngun thc: N1,N3,N5,N7; bn ngun o: N2,N4,N6,N8 (h.69).

(Ox,Oy), (Oz,Ot) trng lp loi ba. Bn ngun thc: N1,N4,N5,N8; bn ngun o: N2,N3,N6,N7 (h.70).

(Ox,Oy), (Oz,Ot) trng lp loi bn. Hai ngun thc: N1,N5; su ngun o: N2,N3,N4,N6,N7,N8 (h.71).

Vi i = 1,2,3, . . . ,8, ta k hiu Ci l tp hp cc ct tuyn ca (Ox,Oy),(Oz,Ot) c ngun sinh l Ni.

1) C nm trng hp cn xem xt.Trng hp 1. (Ox,Oy), (Oz,Ot) khng trng lp.V B(Ox,Oy,Oz,Ot) c tm ngun thc: N1,N2,N3,N4,N5,N6,N7,N8;

khng ngun o nn c hai kh nng xy ra.Kh nng 1.1. C1C3C5C7.V , cng du nn, theo ch 144, C1C3C5C7. Do

, hoc cng ngun hoc i ngun hoc cch ngun. Vy, theo ccnh l 163, 164, 166, ta c

(Ox,Oy).(Oz,Ot).(Ox,Oy).(Oz,Ot) 0.Kh nng 1.2. C2C4C6C8.Tng t kh nng 1.1, ta cng c


60 2. Hng ca gc

Trng hp 2. (Ox,Oy), (Oz,Ot) trng lp loi mt.V B(Ox,Oy,Oz,Ot) c su ngun thc: N1,N3,N4,N5,N7,N8; hai ngun

o: N2,N6 nn c hai kh nng xy ra.Kh nng 2.1. C1C3C5C7.Tng t kh nng 1.1, ta cng c

(Ox,Oy).(Oz,Ot).(Ox,Oy).(Oz,Ot) 0.Kh nng 2.2. C4C8.Tng t kh nng 1.1, ta cng c

(Ox,Oy).(Oz,Ot).(Ox,Oy).(Oz,Ot) 0.Trng hp 3. (Ox,Oy), (Oz,Ot) trng lp loi hai.V B(Ox,Oy,Oz,Ot) c bn ngun thc: N1,N3,N5,N7; bn ngun o:

N2,N4,N6,N8 nn ch xy ra mt kh nng C1C3C5C7. Tngt kh nng 1.1, ta cng c

(Ox,Oy).(Oz,Ot).(Ox,Oy).(Oz,Ot) 0.Trng hp 4. (Ox,Oy), (Oz,Ot) trng lp loi ba.V B(Ox,Oy,Oz,Ot) c bn ngun thc: N1,N4,N5,N8; bn ngun o:

N2,N3,N6,N7 nn c hai kh nng xy ra.Kh nng 4.1. C1C5.Tng t kh nng 1.1, ta cng c


(Ox,Oy).(Oz,Ot).(Ox,Oy).(Oz,Ot) 0.Trng hp 5. (Ox,Oy), (Oz,Ot) trng lp loi bn.V B(Ox,Oy,Oz,Ot) c hai ngun thc: N1,N5; su ngun o: N2,N3,N4,

N6,N7,N8 nn ch xy ra mt kh nng C1C5. Tng t kh nng1.1, ta cng c

(Ox,Oy).(Oz,Ot).(Ox,Oy).(Oz,Ot) 0.Tm li ta lun c

(Ox,Oy).(Oz,Ot).(Ox,Oy).(Oz,Ot) 0.2) C nm trng hp cn xem xt.Trng hp 1. (Ox,Oy), (Oz,Ot) khng trng lp.V B(Ox,Oy,Oz,Ot) c tm ngun thc: N1,N2,N3,N4,N5,N6,N7,N8;

khng ngun o nn c hai kh nng xy ra.Kh nng 1.1. C1C3C5C7.


Khng mt tnh tng qut gi s C1. V , tri du nn, theocc ch 140, 144, C2C4C6C8. Ly C8.

V C1 v C8 l hai ngun k nhau nn v l hai ct tuyn kngun. Do , theo nh l 165,

(Ox,Oy).(Oz,Ot).(Ox,Oy).(Oz,Ot) 0.Theo ch 140, 144, , cng du. Do , theo phn 1,

(Ox,Oy).(Oz,Ot).(Ox,Oy).(Oz,Ot) 0.Mt khc v (Ox,Oy), (Oz,Ot) khng trng lp nn, theo ch 129,

(Ox,Oy).(Oz,Ot) 6= 0.Kt hp ba bt ng thc trn, ta c

(Ox,Oy).(Oz,Ot).(Ox,Oy).(Oz,Ot) 0.Kh nng 1.2. C2C4C6C8.Tng t kh nng 1.1, ta cng c

(Ox,Oy).(Oz,Ot).(Ox,Oy).(Oz,Ot) 0.Trng hp 2. (Ox,Oy), (Oz,Ot) trng lp loi mt.B qua trng hp n gin: hoc (Ox,Oy) hoc (Oz,Ot) l gc nh

hng gia hai tia-khng.V B(Ox,Oy,Oz,Ot) c su ngun thc: N1,N3,N4,N5,N7,N8; hai ngun

o: N2,N6 nn c hai kh nng xy ra.Kh nng 2.1. C1C3C5C7.Khng mt tnh tng qut gi s C1. V , tri du nn, theo

ch 144, C4C8. Ly C8.V C1 v C8 l hai ngun k nhau nn v l hai ct tuyn k

ngun. Do , theo nh l 165,

(Ox,Oy).(Oz,Ot).(Ox,Oy).(Oz,Ot) 0.Theo ch 140, , cng du. Do , theo phn 1,

(Ox,Oy).(Oz,Ot).(Ox,Oy).(Oz,Ot) 0.Mt khc v (Ox,Oy), (Oz,Ot) l gc nh hng gia hai tia-khc

khng nn, theo ch 129, (Ox,Oy).(Oz,Ot) 6= 0.Kt hp ba bt ng thc trn, ta c



62 2. Hng ca gc

Trng hp 3. (Ox,Oy), (Oz,Ot) trng lp loi hai.V B(Ox,Oy,Oz,Ot) c bn ngun thc: N1,N3,N5,N7; bn ngun o:

N2,N4,N6,N8 nn, theo cc ch 140, 144, (Ox,Oy), (Oz,Ot) khng thc hai ct tuyn tri du. Vy trng hp 3 khng xy ra.

Trng hp 4. (Ox,Oy), (Oz,Ot) trng lp loi ba.B qua trng hp n gin: hoc (Ox,Oy) hoc (Oz,Ot) l gc nh

hng gia hai tia-khng.V B(Ox,Oy,Oz,Ot) c bn ngun thc: N1,N4,N5,N8; bn ngun o:

N2,N3,N6,N7 nn c hai kh nng xy ra.Kh nng 4.1. C1C5.Tng t kh nng 2.1, ta cng c


(Ox,Oy).(Oz,Ot).(Ox,Oy).(Oz,Ot) 0.Trng hp 5. (Ox,Oy), (Oz,Ot) trng lp loi bn.V B(Ox,Oy,Oz,Ot) c hai ngun thc: N1,N5; su ngun o: N2,N3,N4,

N6,N7,N8 nn, theo ch 140, (Ox,Oy), (Oz,Ot) khng th c hai cttuyn tri du. Vy trng hp 5 khng xy ra.

Tm li ta lun c


nh l 168. Nu , theo th t l ct tuyn dng, ct tuyn m cahai gc nh hng gia hai tia-khc bt (Ox,Oy), (Oz,Ot) th

1) #(Ox,Oy) #(Oz,Ot) khi v ch khi #(Ox,Oy) #(Oz,Ot).2) #(Ox,Oy) #(Oz,Ot) khi v ch khi #(Ox,Oy) #(Oz,Ot).

Chng minh. 1) Cc iu kin sau tng ng.

1) #(Ox,Oy) #(Oz,Ot).2) (Ox,Oy).(Oz,Ot) 0.3) (Ox,Oy).(Oz,Ot) 0.4)

#

(Ox,Oy) #(Oz,Ot).Ch , theo cc nh l 25, 92, 1 2; theo nh l 167, 2 3; theo cc

nh l 26, 92, 3 4.2) Cc iu kin sau tng ng.

10. Cc vn v hng ca hai gc nh hng 63

1) #(Ox,Oy) #(Oz,Ot).2) (Ox,Oy).(Oz,Ot) 0.3) (Ox,Oy).(Oz,Ot) 0.4)

#

(Ox,Oy) #(Oz,Ot).Ch , theo cc nh l 26, 92, 1 2; theo nh l 167, 2 3, theo cc

nh l 25, 92, 3 4.

10. S cng hng, s ngc hng ca hai gc nhhng gia hai tia

10.1. Hai gc nh hng gia hai tia c cng nh.

nh ngha 169. Hai gc nh hng gia hai tia-khc bt (Ox,Oy),(Oz,Ot) c gi l cng hng nu chng tho mn mt trong hai iukin tng ng trong phn 1 ca nh l 168.

biu th (Ox,Oy), (Oz,Ot) cng hng, ta vit (Ox,Oy) (Oz,Ot).nh ngha 170. Hai gc nh hng gia hai tia-khc bt (Ox,Oy),(Oz,Ot) c gi l ngc hng nu chng tho mn mt trong haiiu kin tng ng trong phn 2 ca nh l 168.

biu th (Ox,Oy), (Oz,Ot) ngc hng, ta vit (Ox,Oy) (Oz,Ot).nh l 171. Gc nh hng gia hai tia-khng cng hng vi migc nh hng gia hai tia-khc bt v cng nh vi n.


nh l 172. Gc nh hng gia hai tia-khng ngc hng vi migc nh hng gia hai tia-khc bt v cng nh vi n.


nh l 173. Vi hai tia Ox,Oy khng i nhau, ta c1) (Ox,Oy) (Ox,Oy).2) (Ox,Oy) (Ox,Oy).3) (Ox,Oy) (Ox,Oy).4) (Ox,Oy) (Ox,Oy).5) (Ox,Oy) (Oy,Ox).

nh l 173 l h qu trc tip ca cc nh l 25, 26, 32.

64 2. Hng ca gc

nh l 174. Nu (Ox,Oy), (Oz,Ot) l hai gc nh hng gia hai tia-khc bt th (Ox,Oy) (Oz,Ot) khi v ch khi (Ox,Oy) (Ot,Oz).


nh l 175. Nu tia Oz hoc thuc min trong hoc trng vi mt tronghai cnh ca gc nh hng gia hai tia-khc bt (Ox,Oy) th

1) (Ox,Oz) (Ox,Oy).2) (Oz,Ox) (Oz,Oy).

nh l 175 l h qu trc tip ca nh l 25, 26, 34.

nh l 176. Vi ba gc nh hng gia hai tia-khc khng v khcbt (Ox,Oy), (Oz,Ot), (Ou,Ov), ta c

1) Nu (Ox,Oy) (Oz,Ot); (Oz,Ot) (Ou,Ov) th (Ox,Oy) (Ou,Ov).2) Nu (Ox,Oy) (Oz,Ot); (Oz,Ot) (Ou,Ov) th (Ox,Oy) (Ou,Ov).3) Nu (Ox,Oy) (Oz,Ot); (Oz,Ot) (Ou,Ov) th (Ox,Oy) (Ou,Ov).

Chng minh. 1) Dng ng thng , ng thi l ct tuyn ca(Ox,Oy), (Oz,Ot), (Ou,Ov). Khng mt tnh tng qut gi s l cttuyn dng ca (Ox,Oy), (Oz,Ot). C hai trng hp cn xem xt.

Trng hp 1. l ct tuyn dng ca (Oz,Ot), (Ou,Ov).V l ct tuyn dng ca (Ox,Oy), (Oz,Ot) v ct tuyn dng

ca (Oz,Ot), (Ou,Ov) nn [Ox,Oy].[Oz,Ot]> 0; [Oz,Ot].[Ou,Ov]> 0.Do [Ox,Oy].[Ou,Ov] > 0. Vy l ct tuyn dng ca (Ox

documents.tips huong trong hinh hoc phang nguyen minh hapdf

Documents