Uchburchaklar tengligi alomatlari

Bu nima va nega kerak

Ikki uchburchak "bir xil" ekanini qanday isbotlash mumkin? Barcha 6 ta elementni (3 tomon, 3 burchak) solishtirish shart emasmi?

Yaxshiyamki, yoʻq. Geometrlar allaqachon isbotlagan: ba'zi hollarda 3 ta element yetarli — agar ular to'g'ri tanlangan bo'lsa. Mana shu "qoida"lar tenglik alomatlari deyiladi.

Amaliy misol: ikki ta yer uchastkasi (uchburchak shaklida) maydoni, perimetri teng ekanini isbotlash uchun barchasini o'lchamay, faqat 3 ta mos o'lchov yetarli bo'ladi.

Asosiy g'oya: tenglik nima

Ta'rif. Agar ikki uchburchakning tomonlari va burchaklari mos-mos teng bo'lsa, bu uchburchaklar teng deyiladi. Belgilash: $\triangle ABC \cong \triangle DEF$.

Tenglik deganda: $A \leftrightarrow D$, $B \leftrightarrow E$, $C \leftrightarrow F$ mos uchlari mos keladi, ya'ni: $AB = DE,\; BC = EF,\; CA = FD,\; \angle A = \angle D,\; \angle B = \angle E,\; \angle C = \angle F.$

Lekin barchasini tekshirmasdan tenglikni qanday aniqlash mumkin? Buning uchun uchta alomat bor.

Uchta tenglik alomati

I-alomat (tomon–burchak–tomon, TBT)

Agar ikki uchburchakda ikki tomon va ular orasidagi burchak mos teng bo'lsa, unda bu uchburchaklar teng.

Sharti: $AB = DE$, $\angle B = \angle E$, $BC = EF$ → $\triangle ABC \cong \triangle DEF$.

Nima uchun ishlaydi? Ikki tomoni va orasidagi burchagi berilgan uchburchak faqat bitta shaklda qurilishi mumkin (kompasni yopib, ikki tomoni ma'lum bo'lgan burchakni o'lchangdek tasavvur qiling).

II-alomat (burchak–tomon–burchak, BTB)

Agar ikki uchburchakda ikki burchak va ular orasidagi tomon mos teng bo'lsa, unda bu uchburchaklar teng.

Sharti: $\angle A = \angle D$, $AB = DE$, $\angle B = \angle E$ → $\triangle ABC \cong \triangle DEF$.

III-alomat (tomon–tomon–tomon, TTT)

Agar ikki uchburchakda uchta tomon mos teng bo'lsa, unda bu uchburchaklar teng.

Sharti: $AB = DE$, $BC = EF$, $CA = FD$ → $\triangle ABC \cong \triangle DEF$.

Nima uchun ishlaydi? Uchta tomon berilsa, uchburchakning shakli to'liq aniqlanadi (uchburchak tengsizligi va qurilish usuli orqali isbotlanadi).

Yechilgan misollar

1-misol. I-alomatni qo'llash

$\triangle ABC$ va $\triangle KLM$ da: $AB = KL = 6$ sm, $BC = LM = 9$ sm, $\angle B = \angle L = 50°$.

Qaysi alomat ishlatiladi?

Ikki mos tomon: $AB = KL$ va $BC = LM$. Ular orasidagi burchak: $\angle B = \angle L$. Bu — I-alomat (TBT).

Xulosa: $\triangle ABC \cong \triangle KLM$. Demak $AC = KM$, $\angle A = \angle K$, $\angle C = \angle M$.

2-misol. II-alomatni qo'llash

$\triangle ABC$ va $\triangle PQR$ da: $\angle A = \angle P = 40°$, $AB = PQ = 7$ sm, $\angle B = \angle Q = 65°$.

Tekshiramiz: $AB = PQ$ — tomon, $\angle A = \angle P$ va $\angle B = \angle Q$ — bu tomonning ikki tomonidagi burchaklar. Bu — II-alomat (BTB).

Xulosa: $\triangle ABC \cong \triangle PQR$. Qolgan elementlar: $BC = QR$, $AC = PR$, $\angle C = \angle R$.

Qo'shimcha: $\angle C = 180° - 40° - 65° = 75°$, ya'ni $\angle R = 75°$ ham.

3-misol. III-alomatni qo'llash

$\triangle ABC$ va $\triangle XYZ$ da: $AB = XY = 5$ sm, $BC = YZ = 8$ sm, $CA = ZX = 6$ sm.

Barcha uch tomon mos teng → III-alomat (TTT).

Xulosa: $\triangle ABC \cong \triangle XYZ$. Barcha burchaklar ham mos teng: $\angle A = \angle X$, $\angle B = \angle Y$, $\angle C = \angle Z$.

4-misol. Tenglikni isbotlash

Isbotlang: $\triangle ABD \cong \triangle CBD$, agar $AB = CB$ va $BD$ — umumiy tomon va $\angle ABD = \angle CBD$.

Qaysi elementlar ma'lum?

• $AB = CB$ (shart bo'yicha)
• $BD = BD$ (umumiy tomon, o'ziga teng)
• $\angle ABD = \angle CBD$ (shart bo'yicha)

$AB = CB$, $\angle ABD = \angle CBD$, $BD = BD$ — I-alomat (TBT): ikki tomon ($AB$, $BD$) va ular orasidagi burchak ($\angle ABD$) mos teng.

Xulosa: $\triangle ABD \cong \triangle CBD$.

5-misol. Qaysi alomat to'g'ri emas?

$\triangle ABC$ va $\triangle DEF$ da: $AB = DE = 5$, $BC = EF = 7$, $\angle A = \angle D = 30°$.

E'tibor bering: $\angle A$ — $AB$ tomoniga tutash burchak, lekin $BC$'ga qarshi burchak. Bu holda burchak ikkala tomonning orasida emas.

Bunday "tomon–tomon–qarshi burchak" sharti tenglikni kafolatlamaydi. Ikki xil uchburchak hosil bo'lishi mumkin.

Shuning uchun I-alomatda burchak aynan ikkala tomonning orasida bo'lishi shart.

Mashq

1. $\triangle ABC$ da $AB=7$, $BC=5$, $\angle B=60°$; $\triangle KLM$ da $KL=7$, $LM=5$, $\angle L=60°$. Bu uchburchaklar tengmi? Qaysi alomat?
2. $\triangle PQR$ da $\angle P=50°$, $PQ=8$, $\angle Q=70°$; $\triangle XYZ$ da $\angle X=50°$, $XY=8$, $\angle Y=70°$. Tenglikni aniqlang.
3. Uchburchaklarning barcha uch tomoni mos teng. Bu qaysi alomat?
4. $\triangle ABC \cong \triangle DEF$ berilgan. Agar $AB=9$, $BC=12$, $\angle C=45°$ bo'lsa, $DE$, $EF$ va $\angle F$ ni toping.
5. $\triangle ABD$ va $\triangle CBD$ da $AD=CD$, $BD$ umumiy tomon. Yana qaysi tenglik berilsa, I-alomat bo'yicha $\triangle ABD = \triangle CBD$ ekanini isbotlash mumkin?
6. Ikki to'g'ri burchakli uchburchakda ikkala katet mos teng. Bu uchburchaklar tengligini qaysi alomat isbotlaydi?
7. $\triangle MNK$ da $MN=NK$. $NL$ — $MK$ ga median. $\triangle MNL \cong \triangle KNL$ ekanini isbotlang.
8. Birinchi uchburchakda $\angle A=70°$, $\angle B=60°$, $AB=10$; ikkinchisida $\angle D=70°$, $\angle E=60°$, $DE=10$. Bu uchburchaklar tengmi?
9. Ikki uchburchakda ikkita mos tomon va bitta mos burchak teng. Bu tenglikni kafolatlaydimi? Nima uchun?
10. $\triangle ABC \cong \triangle CDA$ va $ABCD$ — to'rtburchak. Bu to'rtburchak haqida qanday xulosa chiqarish mumkin?

Javoblar va yechim yo'llari

1. $AB=KL=7$, $BC=LM=5$, $\angle B=\angle L=60°$ — ikki tomon va ular orasidagi burchak. I-alomat (TBT) bo'yicha teng.
2. $\angle P=\angle X=50°$, $PQ=XY=8$, $\angle Q=\angle Y=70°$ — ikki burchak va ular orasidagi tomon. II-alomat (BTB) bo'yicha $\triangle PQR \cong \triangle XYZ$.
3. III-alomat (TTT).
4. $\triangle ABC \cong \triangle DEF$ bo'lgani uchun mos elementlar teng: $DE = AB = 9$, $EF = BC = 12$, $\angle F = \angle C = 45°$.
5. $\angle ADB = \angle CDB$. Bu burchaklar $AD$, $BD$ va $CD$, $BD$ tomonlar orasida turadi, shuning uchun I-alomat (TBT) ishlaydi. (Boshqa yo'l: $AB = CB$ berilsa, III-alomat.)
6. Ikkala katet teng, ular orasidagi burchak esa ikkalasida ham $90°$. Demak I-alomat (TBT) bo'yicha teng.
7. $MN=KN$ (shart), $NL=NL$ (umumiy), $ML=KL$ (median bo'lgani uchun $L$ — $MK$ o'rtasi, ya'ni $ML=KL$). III-alomat bo'yicha $\triangle MNL \cong \triangle KNL$.
8. $AB=DE=10$ tomon va uning ikki yonidagi $\angle A=\angle D=70°$, $\angle B=\angle E=60°$ burchaklar teng. II-alomat (BTB) bo'yicha uchburchaklar teng.
9. Yo'q, kafolatlamaydi. Burchak tomonlar orasida bo'lmasa (5-misoldagidek), ikkita har xil uchburchak hosil bo'lishi mumkin ("ikki yechimli holat").
10. $\triangle ABC \cong \triangle CDA$ bo'lgani uchun $AB=CD$, $BC=DA$ (mos tomonlar). Demak $ABCD$ — parallelogramm (qarama-qarshi tomonlari mos teng).

Keng tarqalgan xatolar

1. Burchakni "orasida" emas, deb ishlatish. ✗ $AB=DE$, $BC=EF$, $\angle A=\angle D$ bo'lsa I-alomat. ✓ I-alomatda burchak ikkala tomon orasida bo'lishi shart: $\angle B = \angle E$ ($AB$ va $BC$ orasidagi).

2. Mos keladigan uchlarni almashtirib yozish. ✗ $\triangle ABC \cong \triangle EDF$ deganda $AB=EF$ deb hisoblash. ✓ Belgilanishda tartib muhim: $A \leftrightarrow E$, $B \leftrightarrow D$, $C \leftrightarrow F$, shuning uchun $AB = ED$.

3. "Uchta burchak teng bo'lsa, uchburchaklar teng" deb o'ylash. ✗ Uch burchak teng bo'lsa ham uchburchaklar teng emas (o'xshash bo'lishi mumkin, lekin o'lchamlari farqli). ✓ Tenglik uchun kamida bitta tomon mos bo'lishi shart.

4. II-alomatda tomonni burchaklar orasida emas deb ishlatish. ✗ $\angle A=\angle D$, $\angle C=\angle F$, $BC=EF$ bo'lsa II-alomat. ✓ II-alomatda tomon $\angle A$ va $\angle B$ orasida — ya'ni $AB$ bo'lishi kerak, $BC$ emas.

5. Tenglik xulosa chiqarilganda barcha elementlarni yozmaslik. ✗ $\triangle ABC \cong \triangle DEF$ dan faqat tomonlarni aytish. ✓ Tenglik uchburchaklarning barcha 6 elementi mos ekanini bildiradi: 3 tomon + 3 burchak.

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Keyingi qadam

Tenglik alomatlari yordamida maxsus uchburchaklarni o'rganamiz — teng yonli uchburchak va uning ajoyib xossalarini.