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Asosiy kontent

Ichki chizilgan burchak teoremasining isboti

Ichki chizilgan burchak bir xil yoyga yopishgan markaziy burchakning yarmiga tengligini isbotlash

Ishga kirishamiz

Teorema isbotiga oʻtishdan oldin, keling, aylanaga oid bir nechta ajoyib atamalarni tushunganimizga ishonch hosil qilib olamiz.
Bu yerda atamalarni mustaqil topa olishingizni tekshirish maqsadida qisqagina mashq berilgan:
Rasmdan foydalanib oʻzgaruvchilarni mos atamalarga toʻgʻrilang.
1

Ajoyib javob! Biz bu terminlardan darsimiz davomida foydalanamiz.

Nimani isbotlamoqchimiz?

Biz aylanaga ichki chizilgan burchak left parenthesis, start color #11accd, \psi, end color #11accd, right parenthesis va markaziy burchak left parenthesis, start color #aa87ff, theta, end color #aa87ff, right parenthesis bitta yoyga tiralsa, qanday ajoyib hodisa roʻy berishini isbotlamoqchimiz: markaziy burchak oʻlchami ichki chizilgan burchak oʻlchamidan ikki marta katta.
start color #aa87ff, theta, end color #aa87ff, equals, 2, start color #11accd, \psi, end color #11accd

Isbotning qisqacha mazmuni

Barcha start color #aa87ff, theta, end color #aa87ff va start color #11accd, \psi, end color #11accd lar uchun (yuqorida taʼkidlaganimizdek) start color #aa87ff, theta, end color #aa87ff, equals, 2, start color #11accd, \psi, end color #11accd ekanini isbotlash maqsadida uchta alohida holatni koʻrib chiqamiz:
A holatB holatC holat
Uchala holat ham markaziy burchak va ichki burchak bir yoyga tiralgan holatdagi barcha mavjud vaziyatlarni koʻrib chiqadi.

A holat: diametr ichki chizilgan burchak start color #11accd, \psi, end color #11accd ning bitta nurida yotadi.

1-qadam: teng yonli uchburchakni toping.

start overline, start color #e84d39, B, C, end color #e84d39, end overline va start overline, start color #e84d39, B, D, end color #e84d39, end overline kesmalar radiuslardir, shu bois ularning uzunligi teng. Bu esa triangle, C, B, D teng yonli ekanini anglatadi, shuningdek, asosiga yopishgan burchaklar ham teng.
m, angle, C, equals, m, angle, D, equals, start color #11accd, \psi, end color #11accd

2-qadam: yoyiq burchakni toping.

angle, start color #e84d39, A, B, C, end color #e84d39 yoyiq burchak, shuning uchun:
θ+mDBC=180mDBC=180θ\begin{aligned} \purpleC \theta + m\angle DBC &= 180^\circ \\\\ m\angle DBC &= 180^\circ - \purpleC \theta \end{aligned}

3-qadam: tenglama tuzing va start color #11accd, \psi, end color #11accd ni toping.

triangle, C, B, D ning ichki burchaklari start color #11accd, \psi, end color #11accd, start color #11accd, \psi, end color #11accd va left parenthesis, 180, degrees, minus, start color #aa87ff, theta, end color #aa87ff, right parenthesis va bizga maʼlumki, har qanday uchburchakning ichki burchaklari yigʻindisi 180, degrees ga teng.
ψ+ψ+(180θ)=1802ψ+180θ=1802ψθ=02ψ=θ\begin{aligned} \blueD{\psi} + \blueD{\psi} + (180^\circ- \purpleC{\theta}) &= 180^\circ \\\\ 2\blueD{\psi} + 180^\circ- \purpleC{\theta} &= 180^\circ \\\\ 2\blueD{\psi}- \purpleC{\theta} &=0 \\\\ 2\blueD{\psi} &=\purpleC{\theta} \end{aligned}
Ajoyib! A holat yuzasidan isbotimizni tugatdik. Atigi ikkita holat qoldi!

B holat: diametr ichki chizilgan burchak start color #11accd, \psi, end color #11accd nurlari oʻrtasida joylashadi.

1-bosqich: bilimingizni ishga solib diametr chizing.

Diametrdan foydalanib start color #11accd, \psi, end color #11accd ni start color #11accd, \psi, start subscript, 1, end subscript, end color #11accd va start color #11accd, \psi, start subscript, 2, end subscript, end color #11accd larga hamda start color #aa87ff, theta, end color #aa87ff ni start color #aa87ff, theta, start subscript, 1, end subscript, end color #aa87ff va start color #aa87ff, theta, start subscript, 2, end subscript, end color #aa87ff larga quyidagicha boʻlamiz:

2-bosqich: ikkita tenglama hosil qilish uchun A holatda oʻrganganlarimizdan foydalaning.

Yangi diagrammada diametr aylanani teng ikkiga boʻladi. Har qaysi yarim aylanada ichki chizilgan burchak boʻlib, nurlarining biri aylana diametridir. Bu A holat bilan bir xil vaziyat, shu bois biz quyidagilarni bilishimiz mumkin:
left parenthesis, 1, right parenthesis, start color #aa87ff, theta, start subscript, 1, end subscript, end color #aa87ff, equals, 2, start color #11accd, \psi, start subscript, 1, end subscript, end color #11accd
va
left parenthesis, 2, right parenthesis, start color #aa87ff, theta, start subscript, 2, end subscript, end color #aa87ff, equals, 2, start color #11accd, \psi, start subscript, 2, end subscript, end color #11accd
A holatda oʻrganganlarimizga muvofiq.

3-bosqich: tenglamani yeching.

θ1+θ2=2ψ1+2ψ2 Qoʻshing (1) va (2)(θ1+θ2)=2(ψ1+ψ2)Oʻzgaruvchilarni guruhlangθ=2ψθ=θ1+θ2 va ψ=ψ1+ψ2\begin{aligned} \purpleC{\theta_1} + \purpleC{\theta_2} &= 2\blueD{\psi_1}+2\blueD{\psi_2}&\small \text{ Qoʻshing (1) va (2)} \\\\\\ (\purpleC{\theta_1} + \purpleC{\theta_2}) &= 2(\blueD{\psi_1}+\blueD{\psi_2}) &\small \text{Oʻzgaruvchilarni guruhlang} \\\\\\ \purpleC{\theta} &= 2\blueD{\psi} &\small\purpleC{\theta=\theta_1+\theta_2} \text{ va } \blueD{\psi=\psi_1+\psi_2} \end{aligned}
B holat tugadi. Yana bitta holat qoldi!

C holat: diametr ichki chizilgan burchak nurlaridan tashqarida.

1-bosqich: bilimingizni ishga solib diametr chizing.

Diametrdan foydalanib ikkita yangi burchak hosil qilamiz: start color #ed5fa6, theta, start subscript, 2, end subscript, end color #ed5fa6 va start color #e07d10, \psi, start subscript, 2, end subscript, end color #e07d10 quyidagi tarzdadir:

2-bosqich: ikkita tenglama hosil qilish uchun A holatda oʻrganganlarimizdan foydalaning.

B holatdagi kabi A holatda oʻrganganlarimizdan foydalanish maqsadida diagramma hosil qildik. Ushbu diagrammadan quyidagilarni aniqlashimiz mumkin:
left parenthesis, 1, right parenthesis, start color #ed5fa6, theta, start subscript, 2, end subscript, end color #ed5fa6, equals, 2, start color #e07d10, \psi, start subscript, 2, end subscript, end color #e07d10
left parenthesis, 2, right parenthesis, left parenthesis, start color #ed5fa6, theta, start subscript, 2, end subscript, end color #ed5fa6, plus, start color #aa87ff, theta, end color #aa87ff, right parenthesis, equals, 2, left parenthesis, start color #e07d10, \psi, start subscript, 2, end subscript, end color #e07d10, plus, start color #11accd, \psi, end color #11accd, right parenthesis

3-bosqich: oʻrniga qoʻying va soddalashtiring.

(θ2+θ)=2(ψ2+ψ)(2)(2ψ2+θ)=2(ψ2+ψ)θ2=2ψ22ψ2+θ=2ψ2+2ψθ=2ψ\begin{aligned} (\maroonC{\theta_2} + \purpleC{\theta}) &= 2(\goldD{\psi_2} + \blueD{\psi})&\small \text{(2)} \\\\\\ (2\goldD{\psi_2} + \purpleC{\theta})&= 2(\goldD{\psi_2} + \blueD{\psi}) &\small \maroonC{\theta_2}=2\goldD{\psi_2} \\\\\\ 2\goldD{\psi_2}+ \purpleC{\theta} &= 2\goldD{\psi_2} + 2\blueD{\psi} \\\\\\ \purpleC{\theta} &= 2\blueD{\psi} \end{aligned}
Bajarib boʻldik! Hamma holatlarda start color #aa87ff, theta, end color #aa87ff, equals, 2, start color #11accd, \psi, end color #11accd ekanini isbotladik.

Qilgan ishlarimiz haqida qisqacha maʼlumot

Biz markaziy burchak oʻlchami ichki chizilgan burchak oʻlchamidan ikki marta katta ekanini isbotladik, bu yerda har ikki burchak bir yoy ostida kesishadi.
Isbotimizni uch xil holat bilan boshladik. Uchala holat ham markaziy burchak va ichki burchak bir yoy ostida kesishgan holatdagi barcha mavjud vaziyatlarni koʻrib chiqdi.
A holatB holatC holat
A holatda teng yonli uchburchak va yoyiq burchaklarni topdik. Ulardan start color #11accd, \psi, end color #11accd va start color #7854ab, theta, end color #7854ab ishtirokida bir necha tenglamalar tuzdik. Qisqagina algebradan foydalanib start color #aa87ff, theta, end color #aa87ff, equals, 2, start color #11accd, \psi, end color #11accd ligini isbotladik.
B va C holatlarda aql bilan diametrni kiritdik:
B holatC holat
Bu A holatda erishgan natijalarimizdan foydalanish imkonini berdi. B holat va C holatlarda tasvirdagi oʻzgaruvchilardan iborat tenglama hosil qildik, bunga faqat A holatda oʻrganganlarimizga tayandik. Teglamalarimizni tuzib boʻlgach, bir necha algebraviy amallarni bajarib, start color #aa87ff, theta, end color #aa87ff, equals, 2, start color #11accd, \psi, end color #11accd ekanini isbotladik.