Question

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Answer 1

[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]
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