$DE//BC$
$→\widehat{ADE}=\widehat{ABC}$ (đồng vị)
$→\widehat{AED}=\widehat{ACB}$ (đồng vị)
mà $\widehat{ABC}=\widehat{ACB}$ ($ΔABC$ cân tại $A$)
$→\widehat{ADE}=\widehat{AED}$
mà $\widehat{ADE}+\widehat{BDE}=180^o$
$\widehat{AED}+\widehat{CED}=180^o$
$→\widehat{BDE}=\widehat{CED}$