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Pe: 2008-03-14T12:56:41+02:00 In: In:

Geometrie – cls7

Se considera paralelogramul ABCD cu m(BAD)<90gr si punctele M apartine(CB si N apartine (CD astfel incat
m(MAB)=m(BAC) si m(NAD)=m(DAC). Demonstrati ca daca AM=AN atunci MN parallel BD.

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  1. user (0)

    1.

        \[ \begin{array}{l} {\rm Consideram AN = AM} \\ {\rm In triunghiul CAN, AD = bisectoare } \Rightarrow \frac{{{\rm CD}}}{{ND}} = \frac{{AC}}{{AN}} \\ {\rm In triunghiul AMC, AB = bisectoare} \Rightarrow \frac{{{\rm MB}}}{{{\rm BC}}} = \frac{{AM}}{{AC}} \Rightarrow \frac{{BC}}{{MB}} = \frac{{AC}}{{AM}} \\ \Rightarrow \frac{{CD}}{{ND}} = \frac{{{\rm BC}}}{{{\rm MB}}} \Rightarrow ({\rm teorema lui Thales)BD||MN} \\ \end{array} \]

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  2. user (0)

    Invers (stiind ca MN||BD, demonstrati ca AN=AM) incearca sa o faci tu 🙂

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