Giải:
a, Ta có: \(\widehat{ADE}=\widehat{BAD}\) ( so le trong do ED // AB )
Mà \(\widehat{BAD}=\widehat{CAD}\left(gt\right)\)
\(\Rightarrow\widehat{ADE}=\widehat{CAD}\)
\(\Rightarrowđpcm\)
b, Nối F với D
Xét \(\Delta BFD,\Delta EDF\) có:
\(\widehat{BFD}=\widehat{EDF}\) ( so le trong do EF // BC )
FD: cạnh chung
\(\widehat{BDF}=\widehat{EFD}\) ( so le trong do EF // BC )
\(\Rightarrow\Delta BFD=\Delta EDF\left(g-c-g\right)\)
\(\Rightarrow\widehat{FBD}=\widehat{DEF}\) ( góc t/ứng )
hay \(\widehat{ABC}=\widehat{DEF}\)
\(\Rightarrowđpcm\)