a)Ta có:
△ABC cân tại A
\(\Rightarrow\widehat{ABC}=\widehat{ACB}\)
\(\Rightarrow\frac{1}{2}\widehat{ABC}=\frac{1}{2}\widehat{ACB}\)
\(\Rightarrow\widehat{ABD}=\widehat{DBC}=\widehat{ACE}=\widehat{ECB}\)
Xét △ABD và △ACE có:
\(\widehat{A}\) chung
AB=AC (gt)
\(\widehat{ABD}=\widehat{ACE}\left(cmt\right)\)
⇒△ABD =△ACE (gcg)
b)Xét △EBC và △DCB có:
\(\widehat{EBC}=\widehat{DCB}\left(gt\right)\)
BC chung
\(\widehat{BCE}=\widehat{CBD}\)(cmt)
⇒△EBC = △DCB (gcg)