A. ED ll AC =>(TEOREMA LUI THALES)[tex] \frac{BD}{AB}= \frac{EB}{BC}<= >\frac{4}{10}= \frac{EB}{15}=> EB = 6 [/tex]
AD=AB-BD=6 cm
EC=BC-EB=5 cm
B.
DE ll AC=> (TEORMEA LUI THALES) =>[tex] \frac{BD}{AD} =\frac{BE}{EC} => \frac{BD}{6} = \frac{12}{3} => BD=24[/tex]
AB=BD+AD=30
BC=BE+EC=15
C.BD=AB-AD=14
DE llAC => (TEOREMA LUI THALES) [tex] \frac{AB}{BD} = \frac{BC}{EB} <=> \frac{24}{14} = \frac{BC}{7} =>BC= \frac{24*7}{14}=12 [/tex]
EC=BC-EB=12-7=5
D.AB=BD+AD=14
DE llAC => (TEOREMA LUI THALES) =>[tex] \frac{AB}{BD} = \frac{BC}{BE} <=> \frac{14}{8} = \frac{BC}{4}=>BC= \frac{14*4}{8} =7
[/tex]
EC=BC-DE=7-4=3
