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In triunghiul isoscel ABC cu m(

Răspuns :

[tex]Varful~este~A~deoarece~unghiurile~de~la~baza~unui~triunghi \\ \\ isoscel~nu~pot~fi~obtuzunghice~(ar~rezulta~ca~suma~masurilor \\ \\ unghiurilor~unui~triunghi~este~mai~mare~de~180 \textdegree, absurd) .\\ \\ m(\ \textless \ ABC)=m(\ \textless \ ACB)= \frac{180 \textdegree - m(\ \textless \ BAC)}{2}= \frac{180 \textdegree-120 \textdegree}{2} =30 \textdegree . \\ \\ m(\ \textless \ BAD)=m(\ \textless \ BAC)-m(\ \textless \ DAC)=120 \textdegree -90 \textdegree =30 \textdegree . \\ \\ Deci~ \Delta DAB-~isoscel \Rightarrow BD=AD. [/tex]

[tex]Fie~O~-mijlocul~lui~[BC] \Rightarrow m(\ \textless \ CAO)= \frac{m(\ \textless \ BAC)}{2}=60 \textdegree. \\ \\ m(\ \textless \ DAO)=m(\ \textless \ DAC)-m(\ \textless \ CAO)=90 \textdegree-60 \textdegree=30 \textdegree. \\ \\ \Delta AOD-dreptunghic~in~O~cu~m(\ \textless \ DAO)=30 \textdegree \Rightarrow DO= \frac{AD}{2}= \frac{BD}{2} . \\ \\ BC= 2 \cdot BO=2(BD+DO)=2(BD+ \frac{BD}{2})=2BD+BD=3BD. \\ \\ \frac{BD}{BC}= \frac{BD}{3BD}= \frac{1}{3}. [/tex]