Vì \(\Delta ABC\) cân tại A (gt)
=> \(\widehat{B}=\widehat{C}\)
mà \(\widehat{BAC}+\widehat{B}+\widehat{C}=180^o\) (t/c \(\Delta\))
=> \(\widehat{B}=\dfrac{180^o-\widehat{BAC}}{2}\)
CMTT, ta có: \(\widehat{D_1}=\dfrac{180^o-\widehat{BAC}}{2}\)
Do đó: \(\widehat{B}=\widehat{D_1}\)
=> BC // DE (dhnb)