A cikin wannan ɗaba'ar, za mu yi la'akari da ɗaya daga cikin manyan ka'idojin Euclidean geometry - Stewart's theorem, wanda ya karɓi irin wannan suna don girmama masanin lissafin Ingilishi M. Stewart, wanda ya tabbatar da hakan. Za mu kuma yi nazari dalla-dalla misali na warware matsalar don ƙarfafa abin da aka gabatar.
Bayanin ka'idar
Dan triangle ABC. Ta gefensa AC batu da aka dauka D, wanda aka haɗa zuwa saman B. Mun yarda da sanarwa mai zuwa:
- AB = a
- BC = ba
- BD = p
- AD = x
- DC = kuma
Don wannan triangle, daidaiton gaskiya ne:
Aikace-aikace na theorem
Daga ka'idar Stewart, ana iya samun dabaru don nemo tsaka-tsaki da bisector na triangle:
1. Tsawon bisector
bari lc aka zana bisector zuwa gefe c, wanda aka raba zuwa sassa x и y. Bari mu dauki sauran bangarorin biyu na triangle a matsayin a и b… A wannan yanayin:
2. Tsawon tsakiya
bari mc shine tsakiyar juya zuwa gefe c. Bari mu nuna sauran bangarorin biyu na triangle a matsayin a и b… Sannan:
Misalin matsala
An ba da triangle ABC A gefe AC daidai da 9 cm, batu da aka dauka D, wanda ke raba gefe don haka AD tsawon sau biyu DC. Tsawon sashin da ke haɗa ƙarshen B da nuna D, 5 cm ne. A wannan yanayin, triangle da aka kafa ABD isosceles ne. Nemo ragowar bangarorin triangle ABC.
Magani
Bari mu kwatanta yanayin matsalar a cikin hanyar zane.
AC = AD + DC = 9cm. AD tsawon DC biyu, watau AD = 2DC.
A sakamakon haka, 2DC + DC = 3DC u9d XNUMX cm. Don haka, DC = 3 cm, AD = 6cm.
Domin triangle ABD - isosceles, da kuma gefe AD yana da 6 cm, don haka suna daidai AB и BDIe AB = 5cm.
Ya rage kawai don nemo BC, samun dabara daga ka'idar Stewart:
Muna maye gurbin sanannun dabi'u cikin wannan furci:
Ta wannan hanyar, BC = √52 ≈ 7,21 cm.