[tex]\it [AM - bisectoare \Rightarrow\ \widehat{MAB} = \widehat{MAC} = \dfrac{m(\hat A)}{2}=\dfrac{80^o}{2}=40^o\\ \\ B'B||AM,\ AB-secant\breve{a} \Rightarrow m(\widehat{ABB'})= m(\widehat{MAB})=40^o\ (alterne\ interne)\\ \\ B'B||AM,\ B'C-secant\breve{a} \Rightarrow m(\widehat{AB'B})= m(\widehat{MAC})=40^o\ (unghiuri\ corespondente)\\ \\ C'C||AM,\ AC-secant\breve{a} \Rightarrow m(\widehat{ACC'})= m(\widehat{MAC})=40^o\ (alterne\ interne)[/tex]
[tex]\it C'C||AM,\ C'B-secant\breve{a} \Rightarrow m(\widehat{AC'C})= m(\widehat{MAB})=40^o\ (unghiuri\ corespondente)[/tex]
