a)
[tex]\it DE||BC \Rightarrow \widehat{DEA} = \widehat{BCA} = 70^o\ (unghiuri\ corespondente)\\ \\ CD-bisectoare \Rightarrow \widehat{DCE}=\dfrac{\widehat{BCA}}{2}=\dfrac{70^o}{2}=35^o\\ \\ \widehat{DEA} \ -\ unghi\ exterior\ pentru\ \ \Delta EDC \Rightarrow \widehat{DEA}=\widehat{DCE}+\widehat{EDC} \Rightarrow\\ \\ \Rightarrow 70^o=35^o+\widehat{EDC}|_{-35^o} \Rightarrow \widehat{EDC}=35^o[/tex]
[tex]\it Din\ \Delta EDC \Rightarrow \widehat{CED} = 180^o-2\cdot35^o=180^o-70^o=110^o.\\ \\ Deci,\ \Delta EDC\ are\ unghiurile\ (35^o,\ \ 35^o,\ \ 110^o)[/tex]
b)
[tex]\it \widehat{EDC}=\widehat{DCE}=35^o \Rightarrow \Delta EDC -isoscel \Rightarrow\ bisectoarea\ EF\ includ e\ \^{i}n\breve al\c{\it t}imea[/tex]
[tex]\it corespunz\breve atoare\ laturii\ DC \Rightarrow EF\perp DC[/tex]
c)
Diagonalele patrulaterului DFCE sunt perpendiculare,
deci DFCE = romb ⇒ DE = DF ⇒ ΔDEF - isoscel
