[tex]\it BD=BC \Rightarrow \Delta BCD-isoscel\ \Rightarrow m(\widehat{BDC})=m(\widehat{BCD})=20^o[/tex]
[tex]\it \widehat{ABC}\ - unghi\ exterior\ triunghiului\ BCD \Rightarrow m(\widehat{ABC})=20^o+20^o=40^o\\ \\ m(\widehat{BCA})=50^o\ (complementul\ lui\ 40^o)[/tex]
