Fales teoremasi

Bu nima va nega kerak

Qadimgi yunon olimi Fales Milatlik Misr ehromlarining balandligini quyosh nuri va soya yordamida aniqlagan: yerga tik qadalgan tayoq soyasi o'z uzunligiga tenglashgan onda ehrom soyasining uzunligini o'lchagan. Quyosh nurlari yerga o'zaro parallel tushgani sababli tayoq va uning soyasi nisbati ulkan ehrom va uning soyasiga teng bo'ladi.

Bu qonuniyat fanga Fales teoremasi va proporsional kesmalar haqidagi teorema nomi bilan kirdi.

Fales teoremasi nima uchun kerak?

• Amaliyotda: Bino, minora yoki daraxt balandligini to'g'ridan-to'g'ri o'lchamasdan aniqlashda;
• Chizmachilikda: Kesmani o'lchovsiz, sirkul va chizg'ich yordamida teng $n$ ta bo'lakka yoki berilgan nisbatda bo'lishda;
• Geometriyada: Uchburchaklar o'xshashligini isbotlashda va DTM masalalarini yechishda asosiy vosita hisoblanadi.

---

Asosiy g'oya va teoremalar

1. Fales teoremasi (teng kesmalar haqida)

[!IMPORTANT] Fales teoremasi. Agar burchakning bir tomonida ketma-ket teng kesmalar ajratilib, ularning uchlari orqali parallel to'g'ri chiziqlar o'tkazilsa, bu chiziqlar burchakning ikkinchi tomonida ham o'zaro teng kesmalar ajratadi.

        O (burchak uchi)
       / \
  A1  /___\  B1   (l1)
     /     \
 A2 /_______\ B2  (l2)
   /         \
A3/___________\B3 (l3)

Agar $OA_1 = A_1A_2 = A_2A_3$ va $A_1B_1 \parallel A_2B_2 \parallel A_3B_3$ bo'lsa: $OB_1 = B_1B_2 = B_2B_3.$

Bu qoida ixtiyoriy ikki to'g'ri chiziqni kesuvchi parallel to'g'ri chiziqlar uchun ham o'rinlidir.

2. Proporsional kesmalar haqidagi teorema

Parallel to'g'ri chiziqlar burchak tomonlarida proporsional kesmalar ajratadi: $\frac{OA_1}{A_1A_2} = \frac{OB_1}{B_1B_2} \quad \text{va} \quad \frac{OA_1}{OA_2} = \frac{OB_1}{OB_2}.$

3. Uchburchakda tomonga parallel to'g'ri chiziq xossasi

$\triangle ABC$ da $AB$ tomonda $M$, $AC$ tomonda $N$ nuqta olinib, $MN \parallel BC$ kesma o'tkazilsa:

1. Yon tomonlardagi kesmalar nisbati: $\frac{AM}{MB} = \frac{AN}{NC}.$
2. Qismning butun tomonga nisbati asosga parallel kesma nisbatiga teng: $\frac{AM}{AB} = \frac{AN}{AC} = \frac{MN}{BC}.$

[!WARNING] Asoslar nisbati butun tomonga nisbatlanadi: $\dfrac{MN}{BC} = \dfrac{AM}{AB}$. $\dfrac{MN}{BC} = \dfrac{AM}{MB}$ deb yozish qo'pol xatodir!

4. Uchburchak bissektrisasi xossasi

[!IMPORTANT] Teorema. Uchburchak bissektrisasi qarama-qarshi tomonni qolgan ikki tomonga proporsional bo'laklarga ajratadi.

$\triangle ABC$ da $AD$ bissektrisa bo'lsa ($D \in BC$): $\frac{BD}{DC} = \frac{AB}{AC}.$

---

Yechilgan misollar

1-misol. Fales teoremasi bo'yicha hisoblash

Burchakning bir tomonida $OA_1 = A_1A_2 = A_2A_3 = A_3A_4$ teng kesmalar ajratilgan. Ularning uchlaridan o'tuvchi parallel to'g'ri chiziqlar ikkinchi tomonni $B_1, B_2, B_3, B_4$ nuqtalarda kesadi. Agar $B_1B_2 = 6\text{ sm}$ bo'lsa, $OB_4$ kesmani toping.

• Fales teoremasiga ko'ra $OB_1 = B_1B_2 = B_2B_3 = B_3B_4 = 6\text{ sm}$.
• Butun kesma: $OB_4 = 4 \cdot 6 = 24\text{ sm}$.

Javob: $24\text{ sm}$.

2-misol. Uchburchakda proporsional kesmalar

$\triangle ABC$ da $MN \parallel BC$ ($M \in AB, N \in AC$). Agar $AM = 6\text{ sm}$, $MB = 9\text{ sm}$, $AN = 8\text{ sm}$ bo'lsa, $NC$ kesma va $AC$ tomon uzunligini toping.

• Proporsional kesmalar teoremasi bo'yicha: $\dfrac{6}{9} = \dfrac{8}{NC}$.
• $NC = \dfrac{9 \cdot 8}{6} = 12\text{ sm}$.
• $AC = AN + NC = 8 + 12 = 20\text{ sm}$.

Javob: $NC = 12\text{ sm}$, $AC = 20\text{ sm}$.

3-misol. Trapetsiyada asoslarga parallel kesma

$ABCD$ trapetsiyada asoslar $BC = 10\text{ sm}$, $AD = 25\text{ sm}$. Yon tomonlaridagi $P \in AB$ va $Q \in CD$ nuqtalar orqali $PQ \parallel AD$ kesma o'tkazilgan. Agar $AP : PB = 3 : 2$ bo'lsa, $PQ$ kesma uzunligini toping.

• $B$ uchidan $CD$ ga parallel $BK$ o'tkazamiz ($K \in AD$), u $PQ$ ni $T$ nuqtada kessin.
• $TQ = KD = BC = 10\text{ sm}$ va $AK = 25 - 10 = 15\text{ sm}$.
• $\triangle ABK$ da $PT \parallel AK$. $AP : PB = 3 : 2 \implies PB = 2x, BA = 5x$: $\dfrac{PT}{15} = \dfrac{2}{5} \implies PT = 6\text{ sm}$.
• $PQ = PT + TQ = 6 + 10 = 16\text{ sm}$.

Javob: $16\text{ sm}$.

4-misol. Uchburchak bissektrisasi xossasi

$\triangle ABC$ da $AB = 12\text{ sm}$, $AC = 18\text{ sm}$, $BC = 20\text{ sm}$. $A$ burchak bissektrisasi $BC$ tomonni $D$ nuqtada kesadi. $BD$ va $DC$ kesmalarni toping.

• Bissektrisa xossasi bo'yicha: $\dfrac{BD}{DC} = \dfrac{12}{18} = \dfrac{2}{3}$.
• $BD = 2x, DC = 3x \implies 2x + 3x = 20 \implies x = 4\text{ sm}$.
• $BD = 2 \cdot 4 = 8\text{ sm}, DC = 3 \cdot 4 = 12\text{ sm}$.

Javob: $BD = 8\text{ sm}$, $DC = 12\text{ sm}$.

---

Mashq

1. Burchak tomonlarini kesuvchi parallel chiziqlar bir tomonda $4\text{ sm}$ va $4\text{ sm}$ li teng kesmalar ajratdi. Ikkinchi tomondagi birinchi kesma $7\text{ sm}$ bo'lsa, ikkinchisini toping.
2. $\triangle ABC$ da $DE \parallel AC$ ($D \in AB, E \in BC$). $BD = 4\text{ sm}$, $DA = 6\text{ sm}$, $BE = 6\text{ sm}$ bo'lsa, $EC$ kesmani toping.
3. Burchak tomonlarini kesuvchi parallel chiziqlar bir tomonda $5\text{ sm}$ va $15\text{ sm}$ kesmalar ajratdi. Ikkinchi tomondagi kesmalar yig'indisi $24\text{ sm}$ bo'lsa, kichigini toping.
4. $\triangle ABC$ da $AB = 15\text{ sm}$, $BC = 12\text{ sm}$, $AC = 18\text{ sm}$. $B$ burchak bissektrisasi $AC$ tomonni $D$ nuqtada kesadi. $AD$ va $DC$ kesmalarni toping.
5. Uzunligi $28\text{ sm}$ bo'lgan kesma Fales teoremasi bilan $3:4$ nisbatda bo'lingan. Katta bo'lak uzunligini toping.
6. $\triangle ABC$ da $MN \parallel AC$ ($M \in AB, N \in BC$). $AM : MB = 2 : 3$ va $AC = 25\text{ sm}$ bo'lsa, $MN$ kesmani toping.
7. $\triangle ABC$ da $AD$ bissektrisa. $BD = 5\text{ sm}$, $DC = 7\text{ sm}$ va perimetri $36\text{ sm}$ bo'lsa, $AB$ va $AC$ tomonlarni toping.
8. To'g'ri burchakli $\triangle ABC$ da ($\angle C = 90^\circ$) $AC = 12\text{ sm}$, $BC = 16\text{ sm}$, $AB = 20\text{ sm}$. $A$ burchak bissektrisasi $BC$ katetni $K$ nuqtada kessa, $BK$ kesmani toping.

Javoblar va yechim yo'llari

1. Fales teoremasiga ko'ra ikkinchi tomonda ham kesmalar teng: $7\text{ sm}$. Javob: 7 sm.
2. $\dfrac{4}{6} = \dfrac{6}{EC} \implies EC = \dfrac{36}{4} = 9\text{ sm}$. Javob: 9 sm.
3. Nisbat $5 : 15 = 1 : 3$. Kesmalar $x$ va $3x$. $4x = 24 \implies x = 6\text{ sm}$. Javob: 6 sm.
4. $\dfrac{AD}{DC} = \dfrac{15}{12} = \dfrac{5}{4}$. $9x = 18 \implies x = 2\text{ sm}$. $AD = 10\text{ sm}$, $DC = 8\text{ sm}$. Javob: AD = 10 sm, DC = 8 sm.
5. $3x + 4x = 28 \implies 7x = 28 \implies x = 4\text{ sm}$. Katta bo'lak: $4 \cdot 4 = 16\text{ sm}$. Javob: 16 sm.
6. $MB = 3x, AB = 5x$. $\dfrac{MN}{25} = \dfrac{3}{5} \implies MN = 15\text{ sm}$. Javob: 15 sm.
7. $\dfrac{AB}{AC} = \dfrac{5}{7} \implies AB = 5x, AC = 7x$. $BC = 12\text{ sm}$. $12x + 12 = 36 \implies x = 2\text{ sm}$. $AB = 10\text{ sm}, AC = 14\text{ sm}$. Javob: AB = 10 sm, AC = 14 sm.
8. $\dfrac{CK}{KB} = \dfrac{12}{20} = \dfrac{3}{5}$. $8x = 16 \implies x = 2\text{ sm}$. $BK = 5 \cdot 2 = 10\text{ sm}$. Javob: 10 sm.

---

Keng tarqalgan xatolar

1. Yon kesmalar nisbati bilan asoslar nisbatini adashtirish. ✗ $\triangle ABC$ da $MN \parallel BC$ bo'lganda $\dfrac{MN}{BC} = \dfrac{AM}{MB}$ deb hisoblash. ✓ Asosga parallel kesma butun tomonga nisbatan olinadi: $\dfrac{MN}{BC} = \dfrac{AM}{AB}$.

2. Bissektrisa doimo tomonni teng ikkiga bo'ladi deb o'ylash. ✗ Ixtiyoriy uchburchakda bissektrisa qarama-qarshi tomonni teng ikkiga bo'ladi. ✓ Bissektrisa faqat teng yonli uchburchak asosida medianaga teng. Boshqa hollarda u tomonni yon tomonlarga proporsional bo'laklarga ajratadi: $\dfrac{BD}{DC} = \dfrac{AB}{AC}$.

3. Parallel bo'lmagan to'g'ri chiziqlarga Fales teoremasini qo'llash. ✗ Chiziqlar parallel bo'lmasa ham kesmalar nisbatini tenglashtirish. ✓ Fales teoremasi faqat kesuvchi chiziqlar o'zaro parallel bo'lgandagina o'rinlidir.

4. Proporsiyada tartibni buzish. ✗ $\dfrac{AM}{MB} = \dfrac{NC}{AN}$ ko'rinishida yozish. ✓ Mos nuqtalar tartibiga rioya qilish shart: $\dfrac{AM}{MB} = \dfrac{AN}{NC}$.

---

DTM uslubidagi test

Ushbu mavzuni mustahkamlash uchun berilgan 5 ta test savolini ishlab, javoblaringizni tekshiring.

Keyingi qadam

Fales teoremasi va proporsional kesmalar xossalarini to'liq egalladik. Endi ushbu fundamental bilimlar asosida geometriyadagi eng muhim mavzulardan biri — uchburchaklar o'xshashlik alomatlari mavzusiga o'tamiz.