INTERNATIONAL JOURNAL OF SCIENTIFIC RESEARCHERS
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CHEVA TEOREMASINI ISBOTLASH
A.Kalandarov, M.Anorbayev
Guliston davlat universiteti matematika kafedrasi o’qituvchisi
Teorema.(Cheva)
Agar ABC uchburchakda AK, BN, va CM chevianalar konkurent bo`lsa, u
holda
munosabat o`rinli bo`ladi.
Isbot
. Bizga ABC uchburchak berilgan bo`lsin.
1-chizma
1) Biz teoremani isbot qilish uchun dastlab BN kesmamizga parallel qilib M nuqtadan AC
tomonga MF kesmani chizib olamiz. BN||MF
2)Fales teoremasiga ko`ra: burchak <CAB uchun hosil bo`ladigan ifodalarni yozib olamiz.Avval
Fales teoremasi haqida ma`lumotlar keltirib o`tamiz.
INTERNATIONAL JOURNAL OF SCIENTIFIC RESEARCHERS
ISSN: 3030-332X Impact factor: 8,293
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Fales teoremasi.
Ma`lum bir burchak tomonlarini to`g`ri chiziqlar bilan kesishimiz tufayli mos
ravishda burchak tomonlaridan nisbatlari bir-biriga teng bo`lgan kesmalar ajraladi, ya`ni biz
hozir <CAB burchakni MF va CN parallel to`g`ri chiziqlar bilan kesganimizda mos ravishda AC
tomondan c va d kesmalar ajratildi, AB tomondan esa a va b kesmalar ajratildi.Teoremaga ko`ra :
tenglikka ega bo`lamiz. AF kesmani umumiy a kesma orqali ifodalaydigan bo`lsak:
AF= a – FN
Endi yuqorida hosil bo`lgan tengligimiz yordamida FN kesmaning qiymatini hisoblab olamiz.
3) Burchak <MBA uchun: Fales teoremasi yordamida hosil bo`ladigan ifodalarni yozib olamiz.
<MBA burchagimiz MF va CN parallel to`g`ri chiziqlar yordamida burchak tomonlarini
kesishimiz natijasida BM tomonnni x va y kesmalarga ajratmoqda, BA tomonni b va FN
kesmalarga ajratmoqda.Demak Fales teoremasi bo`yicha yozib oladigan bo`lsak:
tenglikka ega bo`lamiz.Endi tengligimizning chap va o‘ng tomonlarida turgan ifodalarning bir-
biriga nisbatini oladigan bo`lsak doimo ularning nisbati birga teng bo‘ladi, chunki ikkalasi bir-
biriga teng ifodalar .Ikkita bir xil ifodani bir-biriga bo‘ladigan bo‘lsak doimo javobimiz birga
teng bo`ladi.
Demak, bizda hosil bo`lgan bu tenglik yoki formulaga Menelay teoremasinig ifodasi deyiladi.
4) AK kesmamizga parallel qilib NH kesmamizni chizib olamiz. AK || MH.
Fales teoremasiga ko‘ra: burchak <ACB uchun Fales teoremasi hosil qiladiga ifodalarni yozib
olamiz. <ACB Burchak tomonlarini NH va AK parallel to‘g‘ri chiziqlar bilan kesganimizda
quyidagi tenglik hosil bo‘ladi:
INTERNATIONAL JOURNAL OF SCIENTIFIC RESEARCHERS
ISSN: 3030-332X Impact factor: 8,293
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Bunda CH kesmamiz uzunligini m kesma orqali ifodalaydiga bo‘lsak CH=m-HK ga teng bo‘ladi.
Hosil bo‘lgan tenglikni yuqoridagi ifodaga qo‘yadigan bo‘lsak
5) burchak <CBM uchun Fales teoremasi hosil qiladigan ifodalarni yozib chiqamiz: burchak
<CBM tomonlarini MH hamda AK parallel to‘g‘ri chiziqlarimiz n hamda HK kesmalarga, x va y
kesmalarga ajratmoqda.
Tengligimizdagi HK ni o‘rniga yuqoridagi qiymatini qo‘ysak quyidagi ifoda hosil bo‘ladi.
Endi tenglikning ikki tarafini bir-biriga nisbatini olsak:
ifoda hosil bo‘ldi. Hosil bo‘lgan bu ifoda Menelay teoremasing ifodasi hisoblanadi.Demak, bizda
hosil bo‘lgan Menelay teoremasing ikkita ifodalarining ham qiymati birga teng bo‘ldi, bundan
kelib chiqadiki bu ifodalarning ham nisbati birga teng ekan.
INTERNATIONAL JOURNAL OF SCIENTIFIC RESEARCHERS
ISSN: 3030-332X Impact factor: 8,293
Volume 11, issue 2, May 2025
https://wordlyknowledge.uz/index.php/IJSR
worldly knowledge
Index:
google scholar, research gate, research bib, zenodo, open aire.
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37
Hosil bo‘lgan bu ifodaga Cheva teoremasing ifodasi yoki formulasi deyiladi.
FOYDALANILGAN ADABIYOTLAR RO‘YHATI:
1.
Latipov X, Tojiyev Sh, Rustamov R Analitik geometriya va chiziqli algebra. Toshkent.
“o‘qituvchi” 1993 yil.
2.
Ефимов Н. В. Высше геометрия. М. << Наука >> 1971 г.
3.
Dadajonov N.D, Yunusmetov R, Abdullayev T. Geometriya. 2-qisim Toshkent 1989 yil.
