VARIANTA 1:
[tex] \frac{AE}{BE} = \frac{5}{11} => \frac{AE+BE}{BE}= \frac{5+11}{11} <=> \frac{AB}{BE}= \frac{16}{11} [/tex] [tex]=> BE= \frac{11AB}{16} [/tex]
[tex] \frac{AF}{BF}= \frac{3}{5} => \frac{AF+BF}{BF}= \frac{5+3}{5} <=> \frac{AB}{BF}= \frac{8}{5} => BF= \frac{5AB}{8} [/tex]
[tex] \frac{BF}{BE} = \frac{ \frac{5AB}{8} }{ \frac{11AB}{16} }= \frac{5AB}{8}* \frac{16}{11AB} = \frac{10}{11} [/tex]
[tex] \frac{EF}{BE} = \frac{BE-BF}{BE}= 1- \frac{BF}{BE}= 1-\frac{10}{11}= \frac{1}{11} [/tex]
VARIANTA 2:
[tex] \frac{AE}{BE}= \frac{5}{11}=>AE=5a; BE=11a [/tex]
[tex] \frac{AF}{BF}= \frac{3}{5} => AF=3b; BF=5b [/tex]
[tex]AB=AE+BE=5a+11a=16a \\ AB=AF+BF=3b+5b=8b
[/tex]
Din cele doua relatii de mai sus, rezulta: 16a=8b => 2a=b.
EF=BE-BF=11a-5b=11a-10a=a
[tex] \frac{EF}{BE}= \frac{a}{11a}= \frac{1}{11} [/tex]
