Contents
A cikin wannan ɗaba'ar, za mu yi la'akari da ɗaya daga cikin ka'idodin gargajiya na affine geometry - ka'idar Ceva, wanda ya karɓi irin wannan suna don girmama injiniyan Italiyanci Giovanni Ceva. Za mu kuma yi nazarin misali na warware matsalar don ƙarfafa abin da aka gabatar.
Bayanin ka'idar
An ba da triangle ABC, wanda a cikinsa an haɗa kowane juzu'i zuwa ma'ana a gefe.
Don haka, muna samun sassa uku (AA', BB' и CC'), wanda ake kira cevians.
Waɗannan ɓangarorin suna haɗuwa a lokaci ɗaya idan kuma kawai idan daidaiton mai zuwa ya riƙe:
|KUMA'| |BA'| |CB'| = |BC'| |SHIFT'| |AB'|
Hakanan za'a iya gabatar da ka'idar a cikin wannan nau'i (an ƙayyade a cikin wane rabo maki ya raba bangarorin):
Ka'idar trigonometric na Ceva
Lura: duk kusurwoyi sun daidaita.
Misalin matsala
An ba da triangle ABC tare da dige-dige TO', B' и VS' a bangarorin BC, AC и AB, bi da bi. Matsakaicin triangle an haɗa su zuwa wuraren da aka ba su, kuma sassan da aka kafa sun wuce ta hanyar aya ɗaya. A lokaci guda, maki TO' и B' dauka a tsakiyar maki na daidai kishiyar bangarorin. Nemo a cikin wane rabo batu VS' ya raba gefe AB.
Magani
Bari mu zana zane bisa ga yanayin matsalar. Domin saukaka mana, mun yi amfani da bayanin kula mai zuwa:
- AB' = B'C = a
- BA' = A'C = b
Ya rage kawai don tsara rabon sassan bisa ga ka'idar Ceva kuma musanya abin da aka karɓa a ciki:
Bayan rage ɓangarorin, muna samun:
Saboda haka, AC' = C'B, watau batu VS' ya raba gefe AB a cikin rabin
Saboda haka, a cikin triangle, sassan AA', BB' и CC' masu tsaka-tsaki ne. Bayan mun magance matsalar, mun tabbatar da cewa sun haɗu a lokaci ɗaya (mai inganci ga kowane triangle).
lura: ta amfani da ka'idar Ceva, mutum zai iya tabbatar da cewa a cikin triangle a lokaci ɗaya, bisector ko tsayi kuma suna haɗuwa.