[latex]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} eq frac{AD}{AC} =>DE~nu~este~paralel~cu~BC.[/latex] [latex]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} eq frac{AE}{AC}=>DE~nu~este~paralel~cu~BC. [/latex] [latex]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.[/latex] Figurile sunt in imaginea alaturata:)