[img]http://latex.codecogs.com/gif.latex?\dpi{100}&space;ABC[/img][code] [/code]гурвлжны [img]http://latex.codecogs.com/gif.latex?\dpi{100}&space;AB,&space;BC,&space;CA[/img] талууд ба тэдгээрийн үргэлжлэл дээр харгалзан [img]http://latex.codecogs.com/gif.latex?\dpi{100}&space;C_{1},A_{1},B_{1}[/img] цэгүүд нэг шулуун дээр оршиж байх зайлшгүй бөгөөд хүрэлцээтэй нөхцөл [img]http://latex.codecogs.com/gif.latex?\dpi{100}&space;\frac{\left&space;|AB_{1}&space;\right&space;|}{\left&space;|B_{1}C&space;\right&space;|}\cdot\frac{\left&space;|CA_{1}&space;\right&space;|}{\left&space;|A_{1}B&space;\right&space;|}\cdot&space;\frac{\left&space;|BC_{1}&space;\right&space;|}{\left&space;|C_{1}A&space;\right&space;|}=1[/img] байх явдал юм. [img width=540,height=324]http://data.sur.mn/5e9cb87a-da00-42eb-9d83-1468c18fda82.png[/img][br][url=http://www.math-bu.sur.mn/view.aspx?mate=4f5eb0e5-4573-434c-bf56-695c8e49a643][b][br][b]Баталгаа.[br][br][/b] Зураг-1.  [br][img]http://latex.codecogs.com/gif.latex?\dpi{100}&space;\frac{\left&space;|AB_{1}&space;\right&space;|}{\left&space;|B_{1}C&space;\right&space;|}=\frac{m}{n}&space;(1)[/img][br][img width=555,height=338]http://data.sur.mn/6ac5d312-1dfd-466b-8f12-ada645b4b9c1.png[/img][br][br]Зураг-2.  [img]http://latex.codecogs.com/gif.latex?\dpi{100}&space;\frac{\left&space;|CA_{1}&space;\right&space;|}{\left&space;|A_{1}B&space;\right&space;|}=\frac{n}{l}&space;(2)[/img][br][img width=555,height=340]http://data.sur.mn/86aafbfa-96eb-41d9-b73f-faba3426e039.png[/img][br][br]Зураг-3.  [img]http://latex.codecogs.com/gif.latex?\dpi{100}&space;\frac{\left&space;|BC_{1}&space;\right&space;|}{\left&space;|C_{1}A&space;\right&space;|}=\frac{l}{m}&space;(3)[/img][br][br][br][img width=555,height=488]http://data.sur.mn/0cb03d93-fb09-49fe-8a3b-20c024ce0543.png[/img][br][br][br]харьцаа  [img]http://latex.codecogs.com/gif.latex?\dpi{100}&space;(1),(2),(3)[/img]  ëсоор  [img]http://latex.codecogs.com/gif.latex?\dpi{100}&space;\frac{\left&space;|AB_{1}&space;\right&space;|}{\left&space;|B_{1}C&space;\right&space;|}\cdot\frac{\left&space;|CA_{1}&space;\right&space;|}{\left&space;|A_{1}B&space;\right&space;|}\cdot&space;\frac{\left&space;|BC_{1}&space;\right&space;|}{\left&space;|C_{1}A&space;\right&space;|}=\frac{m}{m}\cdot&space;\frac{n}{l}\cdot&space;\frac{l}{m}=1[/img] болж теорем батлагдав.[/b][/url][br]