All Geometri

May 4, 2019 | Author: anii88 | Category: N/A
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Lanjutan BAB III “Apakah ada segitiga yang sama dalam geometri Lobachevsky?” Lobachevsky ?”

1.1

Teorema 4:

“ Dua segitiga kongruen, jika sudut -sudut yang bersesuaian sama”. A

A’

B”

C’

B’

C”

B

C

Gambar. 3.7

Bukti: Anggap teorema ini salah. Maka pasti ada dua segitiga yaitu ΔABC dan Δ A‟B‟C‟ Э =



B‟, dan

C

=



A

=  A‟,

B

C‟, tetapi segitiga tersebut tidak kongruen.

 AB   A' B' (jika tidak, segitiga tersebut kongruen melalui sudut sisi sudut) Maka,  AB

Demikian juga pada  AC   AC    A' C '   dan  BC   BC    B' C ' . Perhatikan tiga segmen  AB ,  AC  ,

 BC  dan  A' B' ,  A C  ,  B C  '

'

'

'

Dari ketiga segmen tersebut ada dua segmen yang lebih besar dari dua segmen yang bersesuaian dari ketiga segmen lain. Konsekuensinya,  AB >  A' B' dan

 AC  >  A' C '

Jadi, dapat ditemukan B” pada

 AB

dan C” pada

 AC 

Э  A' B' =  AB" dan

 A' C '  =  AC " .

Konsekuensinya, ΔA‟B‟C‟  ΔAB”C”

AB”C” =  B‟ =  B Karena  BB”C” merupakan sudut pelurus dari

Sehingga,



 juga merupakan sudut pelurus dari

 B.



B”, maka



BB”C”

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Demikian juga,



B”C”C merupakan sudut pelurus dari



C”, maka



B”C”C juga merupakan sudut pelurus dari  C. Oleh karena itu, segi empat BB”C”C mempunyai jumlah sudut 360°, dimana hal ini kontradiksi dengan Teorema 3 Cololarry 1.

3.5. Teori Daerah Lobachevskian

Mari kita klasifikasikan masalah dengan menguji dasar karakter dari sebuah ukuran bidang untuk segitiga. Perhatikanlah bagaimana bidang tersebut didefinisikan, tentu saja akan mengikuti sifat-sifat: a. Positivity. Untuk masing - masing segitiga mempunyai hubunan unik yang ditentukan oleh bilangan real positif disebut daerah / area.  b. Invariance Under Congruence. Segitiga kongruen mempunyai wilayah yang sama. c. Additivity. Jika segitiga T dibagi menjadi dua segitiga yaitu T 1  dan T2  oleh garis yang ditarik dari titik puncak ke sisi yang dihadapannya, maka wilayah dari T adalah penjumlahan dari T 1 dan T2. Akibatnya beberapa proses untuk pengukuran bidang yang ditentukan oleh sebuah fungsi nilai real didefinisikan untuk semua segitiga yang memenuhi memenuhi a, b, dan c. Ini memberi tahu kita bahwa konsep pengukuran daerah atau daerah fungsi segitiga dengan mengartikan property- property tersebut.

Definisi

Suatu fungsi yang menentukan setiap segitiga dengan spesifikasi bilangan real memenuhi a, b, dan c. Maka fungsi itu disebut sebagai daerah fungsi atau

daerah pengukuran untuk segitiga. Jika μ adalah  suatu fungsi seperti itu dan ABC sebuah segitiga, maka μ (ABC) merupakan nilai dari Δ ABC dan disebut daerah atau ukuran dari ΔABC yang ditentukan oleh μ. Definisi ini tentunya tidak terikat oleh geometri Lobachevskian, namun geometri ini berlaku untuk geometri netral. Kenyataannya dalam geometri

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ini menunjukkan setiap wilayah segitiga merupakan ukuran setengah dari hasil kali alas dan tinggi. Kita lanjutkan dengan mengamati sifat additivity (c), dimana fungsi daerah dapat diperluas sampai bilangan bulat terbatas.

Teorema 5:

Jika setiap segitiga merupakan gabungan dari himpinan terbatas yang tidak  beririsan (1,2,...., n). Maka untuk setiap fungsi daerah μ,

μ (Δ) = μ (1) + μ (2) + .... + μ (n) Definisi: The defect dari segitiga ABC adalah 180 - (  A +

Disini

 A,  B,

dan

C

B

+

 C)

digunakan sebagai derajat pengukuran dari sudut yang

dimaksud, sehingga menghasilkan suatu nilai real, bukan suatu bilngan derajat. Dengan catatan

 A +  B

+

C

< 180.

Teorema 6:

The defect tersebut merupakan fungsi daerah untuk segitiga. Bukti: Sifat (a) mengikuti teorema 3

  A +  B

+

C

< 180°

Sifat (b) segitiga yang kongruen mempunyai sudut-sudut yang bersesuaian sama  besar, sehingga jumlah sudutnya sama dan the defect juga sama. A

B

D Gambar. 3.8

C

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Sifat (c) Diketahui Δ ABC dan D pada BC, AD membagi

Δ ABC menjadi Δ ABD dan Δ

ACD. Jumlah the defect dari kedua segitiga tersebut adalah 180 - (  BAD +

B

+

 BDA)

Dengan mengetahui bahwa

+ 180 - (  CAD +

 BDA +  CDA

C

+

 CDA)

= 180, maka

Jumlah the defect dari kedua segitiga tersebut adalah 180 - (  BAD + 180 - (  A +

 CAD +  B

B

+

+

 C)   BAD +  CAD

 C)  sesuai

dengan definisi diatas.

=

A

maka:

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Lanjutan BAB III

3.6. Riemann's Non-Euclidean Geometri Teori

POSTULAT SEJAJAR RIEMANN

Tidak ada garis-garis sejajar

Teori Riemann tidak hanya memerlukan paralel Euclid dalil tapi dalil-dalil lain  juga. Sebab kita telah menunjukkan, tanpa berasumsi apapun postulat paralel, yang ada garis-garis sejajar (Bab 2, Th. 2, Kor. 3); adanya garis-garis paralel, tidak konsisten dengan dalil-dalil geometri netral. Akibatnya, kita akan menemukan dalil-dalil geometri netral menyiratkan adanya garis-garis parallel. Prosedur alami untuk melakukan ini adalah untuk menganalisis bukti adanya garisgaris paralel (Bab 2, Th.2, Kor. 3) untuk melihat atas mana properti itu tergantung. Melirik bukti, kita melihat bahwa ia mengikuti langsung dari propertiberikut: Properti (A) adalah akibat langsung dari teorema sudut eksterior, jadi kita harus menentukan dalil-dalil teorema sudut eksterior bergantung. Tetapi bukti teorema malaikat eksterior adalah kompleks dan melibatkan penerimaan diam-diam properti grafis untuk dibuang. Namun, ada bukti alternatif properti (A) yang sederhana dan tidak memerlukan dosis malaikat eksterior teorema. Kami menyajikan dan menganalisisnya untuk menurunkan sifat-sifat penting. Teorema

Dua garis tegak lurus terhadap baris yang sama sejajar. L

C

M

M L  N

A

B (a)

 N

B

A

C (b)

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Bukti: Misalkan L sejajar dengan M adalah salah. Kemudian L dan M akan bertemu di titik C (gambar 4,14 (b)). L, M, bertemu dengan N di A, B, masing-masing. 1. Perluas CA panjang sendiri melalui 2.

A ke C‟ Draw C‟B

1. Sebuah segmen dapat dua kali lipat 2. Dua titik menentukan garis

3.  AB  ABC    AB  ABC 

3. SAS

4.  AB  ABC    AB  ABC '

4. Sesuai bagian

Dengan demikian  AB  ABC   juga sudut siku ABC '  adalah sudut siku-siku karena  AB ‟

siku dan BC dan BC  adalah tegak lurus dengan AB. 5. BC dan BC' bertepatan

5. Hanya ada satu garis tegak lurus terhadap baris tertentu  pada suatu titik tertentu dari garis

Jadi AC dan BC atau L dan M memiliki titik C dan C ' 6. Oleh karena itu L dan M bertepatan

6. Dua titik menentukan garis

Ini bertentangan dengan hipotesis kita bahwa L dan M adalah garis yang berbeda. Jadi,  pengandaian kita salah dan teorema berlaku. Jika postulat sejajar Riemann akan dijadikan pegangan, teorema ini harus dipahami. Jadi kita harus membuang (selain postulat paralel Euclid) salah satu prinsip yang digunakan dalam pembuktian. Tentu saja kita ingin mempertahankan sifat-sifat dasar kongruen segitiga dan garis tegak lurus. Dengan pemikiran ini marilah kita menganalisis bukti. Titik penting tampaknya langkah 6, bahwa L dan M serupa karena mereka memiliki poin berbeda C dan C 'yang sama. Langkah ini (dan bukti) akan gagal  jika C dan C 'tidak berbeda, yaitu, jika mereka bersamaan. Bagaimana mereka bisa  bertepatan? Sebaliknya, kita harus bertanya bagaimana kita tahu bahwa mereka berbeda. Ini poin penting dalam pembuktian tidak formal dibenarkan, tetapi tampaknya sudah pasti dari diagram. Dapatkah kita menemukan prinsip geometris untuk membenarkan itu? Untuk menjawab ini, mengamati bahwa secara diam-diam Euclid mengasumsikan  bahwa setiap s etiap garis "memisahkan" bidang menjadi dua sisi yang berlawanan. berlaw anan. Dinyatakan secara lebih tepat: jika L adalah suatu garis, titik-titik bidang, L bukan terletak pada dua  bidang atau rangkaian titik-titik, yang disebut sisi L. sisi ini tidak memiliki titik yang

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C dan C' adalah di sisi N, dan begitu pula poin berbeda. Tanpa pemisahan yang beda C dari C 'tidak memiliki justifikasi formal, dan bukti gagal/salah. Hal ini menunjukkan  bahwa kita dapat membuat sebuah "Riemann" teori geometri dengan membuang dalil  bahwa setiap garis memisahkan bidang. Jika prinsip pemisahan diterima, C dan C 'harus menjadi titik berbeda, tetapi kita masih dapat menghindari kontradiksi pada langkah 6, jika kita meninggalkan prinsip  bahwa dua titik menentukan garis, dan mengizinkan dua garis berpotongan dalam dua titik. Pada pandangan pertama mungkin ini tampaknya pembayaran yang terlalu tinggi, namun itu mengarah pada yang menarik dan bukan teori geometris sederhana.

Ringkasan

Ada dua teori geometris yang mengasumsikan postulat sejajar Riemann. Pertama, setiap dua garis berpotongan dalam tepat satu titik, tetapi tidak ada garis memisahkan  bidang. Kedua, dua garis berpotongan dalam tepat dua titik, dan setiap garis memisahkan  bidang. Teori-teori ini disebut, masing-masing, geometri eliptik tunggal dan geometri eliptik ganda. (Istilah "tunggal" dan "rangkap" mengindikasikan sifat perpotongan dua garis dalam geometri dan istilah "elips" digunakan lebih halus dalam arti klasifikasi  berdasarkan geometri proyektif dimana geometri Euclid dan Lobachevskian disebut  parabola dan hiperbolik).

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BAB IV Teori Geometri Insidensi

4.1.

Teori Dasar Geometri Insidensi

Geometri mengandung: - Unsur-unsur tak terdefinisi - Aksioma - Definisi-definisi - Teorema-teorema Geometri insidensi dapat dikatakan mendasari Geometri Euclides.

Unsur-unsur tak terdefinisi pada sebuah geometri :

- Titik - Garis - Bidang Ketiga unsur dikaitkan satu sama lain dengan sebuah aksioma yaitu system aksioma insidensi.

Ada 6 buah postulat : 1.1 Garis adalah himpunan dari titik-titik yang mengandung paling sedikit dua buah titik. 1.2

Dua buah titik yang berbeda terdapat dalam satu dan hanya satu garis.

1.3

Bidang adalah himpunan titik-titik yang mengandung paling sedikit tiga titik, dimana ketiga titik tersebut tidak terletak pada garis yang sama.

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Definisi

Sebuah himpunan titik-titik bersama dengan himpunan bagian seperti garis dan  bidang yang memenuhi memenuhi postulat 1.1 sampai 1.6 disebut geometri insidensi.

Teorema 1

Dua garis yang berbeda berpotongan pada paling banyak di satu titik. Bukti : Andai g   h,dan  h,dan (g,h) = a,b (hipotesis)

Bukti Karena a,b = (g,h), a,b di g, a,b di h, g berimpit dengan dengan h (postulat 1.2) Dan pernyataan tersebut berlawanan dengan hipotesis jadi haruslah (g,h)

  1

titik

Definisi

 jika a dan b adalah titik-titik yang berbeda, kita gunakan symbol ab untuk menyatakan garis unik yang memuat a dan b, dan disebut garis yang ditetapkan oleh a dan b. dan juga dikatakan garis ab adalah garis yang menghubungkan a dan b (jika a dan b adalah titik yang sama symbol ab tidak terdefinisi).

Definisi

Titik-titik A1, A2, A3,...., An dikatakan segaris atau sejajar, jika ada sebuah garis yang memuat semua titik tersebut. Dengan cara yang sama kita mendefinisikan bentuk/gambar (himp. Dari titik-titik) S1, S2,...., Sn menjadi segaris atau sejajar jika ada sebuah garis yang memuat titik tersebut.

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Bukti B , BC di C, BC Karena A=B, ABC berlawanan dengan yang diketahui ABC. Kesimpulan A B , A, B, C tidak segaris. Andaikan A, B, C segaris segaris A, B,C d¡ g (definisi) Jika B, C di g dan BC, g = BC(aksioma 1.2). Karena A B C segari di g maka  pernyataan ini berlawanan dengan hipotesis hipotesis maka ABC segaris.

Teorema 3

Sebuah garis dan sebuah titik yang tidak terletak pada garis tersebut termuat pada satu  bidang.

Definisi

Andaikan A g. satu-satunya bidang yang memuat g dan ditulis sebagai gA. Andaikan A,B,C berbeda dan tidak segaris. Satu-satunya bidang yang memuat A,B,C ditulis sebagai nbidang ABC.

Definisi:

Dua garis l dan m adalah sejajar apabila l dan m terletak pada bidang yang sama dan tidak mempunyai titik perpotongan.

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Lanjutan BAB IV

4.2. Bidang-bidang Sejajar dan Garis-garis Sejajar Definisi

Dua bidang P dan Q dikatakan sejajar (ditulis P║Q), jika keduanya tidak mempunyai titik temu.

Teorema 6

Jika bidang-bidang P dan Q sejajar, seja jar, dan bidang R berpotongan dengan bidang p dan Q, maka perpotongan R dengan P dan Q merupakan garis-garis yang sejajar.

Bukti: Dengan menggunakan teorema 5 : Jika dua bidang berbeda berpotongan, maka  perpotongannya merupakan merupakan sebuah garis. (i) Akan ditunjukkan bahwa bidang R berbeda dengan bidang P dan Q.

( R ≠ P dan R ≠ Q ) Andaikan R = P. Maka R memotong Q, akibatnya P memotong Q.

Bertentangan dengan P║Q. Pengandaian salah. Jadi, R ≠ P. Andaikan R = Q. Maka R memotong P, akibatnya Q memotong P.

Bertentangan dengan P║Q. Pengandaian salah.

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a di dalam P ( karena L di P ) a di dalam Q ( karena M di Q ) P dan Q berimpit.

Bertentangan dengan hipotesis bahwa P║Q. Selanjutnya L dan M terletak pada bidang yang sama dan tidak berimpit .

Dengan definisi bahwa L║M. Definisi:

Garis-garis L1, L2,

…, Ln 

dikatakan kongkuren, jika garis-garis tersebut

 berpotongan di satu titik. Gambar-gambar S1, S2, ..., S n  dikatakan koplanar, jika ada sebuah bidang yang memuat semua gambar-gambar tersebut.

Teorema 7: Jika tiga garis koplanar secara berpasangan, tetapi semuanya tidak koplanar,

maka ketiga garis tersebut kongkuren atau ketiganya garis tersebut paralel secara berpasangan.

Bukti: Misalkan L, M, dan N tiga buah garis.

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Berarti, P ≠ R. ..............( 3 ) Dari ( 1 ), ( 2 ), dan ( 3 ), menunjukkan bahwa P ≠ Q ≠ R. Berikut ini, bidang-bidang memotong secara berpasangan di garis-garis seperti ditunjukkan pada tabel di bawah ini: Bidang-bidang P, Q Q, R P, R

Garis perpotongan M N L

Andai dua garis bertemu. Katakan L, M bertemu di titik a. Karena a di L. Dari tabel a di P dan di R. Karena a di M. Dari tabel a di P dan di Q. Berarti a di Q dan R.

Dari tabel a di N. Selanjutnya, jika dua dari L, M, N bertemu. Ketiga garis tersebut kongkuren.

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Teorema 9

Tiap bidang memuat 3 garis berbeda yang tidak kongkuren.

Bukti: Dengan postulat 13: Sebarang bidang P memuat 3 titik berbeda yang tidak kolinier a, b, c. Dengan postulat 15:

P memuat ab, bc, ca. dan ab ≠ bc.

Andaikan ab = bc maka c di ab. Berarti a, b, c kolinier. Bertentangan dengan hipotesis.

Jadi, ab ≠ bc. Dengan cara yang sama, ab

≠ ac dan bc ≠ ac.

Sehingga ab, bc, ac adalah garis-garis yang berbeda. Karena ab, bc, ac berpotongan secara berpasangan pada tit ik-titik yang berbeda a, b, c. Titik tersebut tidak dapat menjadi kongkuren.

Corrolary 1

Pada bidang P, jika titik a diberikan. Ada sebuah garis yang tidak termuat. Bukti:

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Teorema 10

Andaikan ada 4 titik a, b, c, d berbeda, tidak kolinier dan tidak koplanar. Maka: (i) Diberikan sebuah bidang, ada sebuah titik tidak di dalam bidang tersebut. (ii) Diberikan sebuah garis, ada sebuah garis menjulur ke garis garis tersebut. (iii) Diberikan sebuah titik, ada sebuah bidang tidak termasuk di titik tersebut. (iv) Ada paling sedikit 6 garis dan paling sedikit 4 bidang.

Bukti: (i) Misalkan diberikan bidang P. Karena a, b, c, d tidak koplanar, paling sedikit satu dari titik tersebut tidak di P. (ii)

Misalkan diberikan garis L. Karena a, b, c, d tidak kolinier, paling sedikit satu dari titik tersebut tidak di L. Misalkan p sebuah titik tidak di L. Perhatikan bidang Lp. Dengan (i) ada sebuah titik g tidak di Lp. Berikut ini garis pq menjulur ke L.

(iii) Misalkan diberikan titik r. Karena a, b, c, dberbeda, ada sebuah titik s berbeda dengan r, Perhatikan garis rs. Dengan (ii) ada garis M menjulur ke rs,

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BAB V Teori Geometri Affin

5.1. Pendahuluan

Teori Geometri Affin merupakan geometri yang berisikan tentang geometri insidensi yang memenuhi postulat sejajar Euclid dalam bentuk Playfair. Geometri insidensi dikatakan geometri Affin jika memenuhi postulat berikut:  Postulat E   Jika titik A tidak terletak terlet ak pada garis l maka terdapat hanya satu  –  satunya  satunya garis m  sedemikian hingga m memuat A dan m // l. Ilustrasi: l . m

.A

Dalam uraian ini, relasi insidensi terhadap titik, garis, dan bidang digunakan notasi dan

kata “pada” (konvensi Veblen dan Young), sebagai berikut:

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Bukti: 

Menurut postulat E , yaitu: Jika titik A tidak terletak pada garis l  maka   maka terdapat dan

hanya satu  –  satunya   satunya garis m sedemikian hingga m memuat A dan m // l . Berarti ada garis tunggal (unik) m sedemikian hingga m pada A dan m // l . 

Menurut definisi garis sejajar, yaitu: dua buah garis adalah sejajar, bila garis itu

terletak pada sebuah bidang dan tidak mempunyai satupun titik persekutuan. Berarti garis l  dan   dan m

sebidang (koplane), kita misalkan pada bidang β. Karena  A pada m, A

pada β, maka A, m pada β 

Menurut teorema 3 bab 7, yaitu : sebuah garis dan sebuah titik yang terletak pada

garis tersebut termuat pada satu bidang. Berarti ada bidang yang memuat titik A dan garis l  

 pada α. Jadi Jadi β = α, sehingga m pada α. Sesuai hipotesis tadi, yaitu: A, l  pada

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Sehingga pengandaian l  berpotongan  berpotongan n menjadi salah. Karena l , n, dan m sebidang, maka menurut definisi garis sejajar artinya l // n.

 , m, n, tidak n, tidak sebidang Kasus 2: l , Ilustrasi:

P l m

n‟ A

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Catatan: Kodireksionalitas garis dapat dianggap sebagai generalisasi dari kesejajaran, karena sebagai tambahan pada kesejajaran. Kodireksionalitas mencakup koinsidensi yang

merupakan jenis kasus “degenerasi” dari kesejajaran. Dalam situasi tertentu lebih mudah mempelajari kodireksionalitas daripada kesejajaran karena sifat formalnya lebih biasa digunakan. Secara khusus kodireksionalitas garis dikatakan merupakan relasi ekivalensi, yakni: Untuk sembarang garis l , m, n maka pernyataan berikut ini berlaku: i.

l  cod  cod l 

ii.

 jika l  cod  cod m, maka m cod l

iii.

 jika l  cod  cod m dan m cod n, maka l  cod  cod n

Perhatikan bahwa (i) dan (ii) tidak berlaku untuk kesejajaran relasi garis, dan (iii)

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5.3. Transversalitas Transversalitas Garis Jika garis l , m koplane (sebidang), maka garis tersebut harus memenuhi salah satu

dari tiga relasi berikut: 1) l  //  // m. 2) l  =  = m, atau 3) l  //  // m dan l ≠ m Dalam kasus ketiga, l  dan  dan m berpotongan dan berbeda. Kasus ini merupakan relasi yang  penting antara dua garis dan sangat berguna berguna dalam studi kesejajaran, dan diperlukan suatu nama. Jadi, kita perkenalkan definisi berikut:

Definisi Kita katakan l  transvers   transvers m, atau l  merupakan   merupakan suatu transversal dari m, atau l dan m adalah transvers, ditulis l  tr  tr m jika l  memotong  memotong m dan l ≠ m.

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Bukti: Misalkan A terletak pada garis l, n. n. Andaikan n tidak transvers m, maka yang terjadi adalah n = m atau n // m. Sehingga haruslah: (i) n ≠ m, maka jika tidak A akan memiliki secara bersama oleh garis sejajar; dan (ii) n // m, maka jika tidak akan ada dua garis berbeda l   dan n, dimana setiap garis tersebut memuat A, dan setiap garis tersebut sejajar dengan m. Hal tersebut kontradiksi dengan postulat E, bahwa mestinya hanya ada satu garis sejaj ar m yang memuat A. Jika pengandaian salah, sehingga n tr m

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Lanjutan BAB V

5.4. Transversalitas Transversalitas Garis dan d an Bidang Definisi

Jika garis l dan bidang α tidak memliki titik sama, dikatakan bahwa l sejajar

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Andaikan terdapat titik lain, yakni C yang sama-sama terletak pada bidang

α dan garis

m, menurut postulat 15, yaitu jika sebuah bidang mengandung dua titik yang sejenis, maka bidang itu memuat garis. Artinya m terletak pada bidang α. Karena m terletak juga pada bidang

β, ini menunjukkan bahwa m ≠n, hal ini

kontradiksi dengan n tr m. Jadi pengandaian salah, dan B satu-satunya titik yang yang

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 Teorema 4 :

Sebuah garis transvers dengan satu dari dua bidang sejajar, maka akan transvers dengan yang laiannya.

Bukti : Diketahui α//β, l tr α. Misalkan A sebuah titik pada β yang tidak pada l, maka menurut postulat E

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BAB VI Teori Urutan Pada Garis

6.1. Konsep Urutan

Urutan adalah salah satu yang paling dasar dari suatu ide matematika. kita

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6.2. Postulat Untuk Keantaraan

Ada banyak system dari postulat betweeness yang dipilih dengn alasan yang sederhana, bukan untuk sulit dipahami dan untuk memfasilitasi pembelajaran urutan dalam bidang dan ruang. Kita mempertimbangkan geometri insidensi secara umum pada pada 11-16 dan mengenal konsep dasar penjumlahan “antara” yang diindikasikan oleh symbol (abc) yang dibaca titik a, b, c adalah urutan abc atau b diantara a dan c. (Postulat E pada

Bab 9

tidak diasumsikan). Kita mengasumsikan mengasumsi kan bahwa relasi “antara”

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B4 sejalan atau memiliki satu dimensi dengan Postulat Pasch (Bab 11) yang

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x€ba hanya jika (bxa)….definisi ba

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