[tex]a)\\AD=6cm\\
BD=12cm\\
AE=5cm\\
ED=10cm\\
~-----/--\\
DE||BC=?\\
-----//- -\\
Aplicam~teorema~lui~Theles:\\\\
\frac{AD}{AB} = \frac{6}{AD+BD} \\\\
\frac{AD}{AB} = \frac{6^(^6}{18} = \frac{1}{3} (1)\\\\
DE=EC=10cm\\\\
\frac{AD}{AC} = \frac{5^(^5}{10} = \frac{1}{5} (2)\\\\
Din~ (1) ~si~(2)=> \frac{AD}{AB} \neq \frac{AD}{AC} =>DE~nu~este~paralel~cu~BC.[/tex]
[tex]b)\\AD=3cm\\
AB=12cm\\
AE=4cm\\
AC=16cm\\
-------/---\\
DE||BC\\
------//----\\
\frac{AD}{AB}= \frac{3^(^3}{12} = \frac{1}{3} (1)\\\\
\frac{AE}{AC} = \frac{4^(^4}{16} = \frac{1}{4} (2)\\\\
Din~(1)~si~(2)=> \frac{AD}{AB} \neq \frac{AE}{AC}=>DE~nu~este~paralel~cu~BC.
[/tex]
[tex]c)\\
BD=5cm\\
AB=25cm\\
CE=6cm\\
AC=30cm\\
------/-----\\
DE||BC=?\\
------//-----\\
BD=AD=5cm\\
CE=AE=6cm\\\\
\frac{AD}{AB} = \frac{5^(^5}{25} = \frac{1}{5} (1)\\\\
\frac{AE}{AC} = \frac{6^(^6}{30} = \frac{1}{5} (2)\\\\
Din~(1)~si~(2)=> \frac{AD}{AB} = \frac{EC}{AC} =>DE||BC.[/tex]
Figurile sunt in imaginea alaturata:)
