a) Vì \(\Delta ABC\) đều (gt)
\(\Rightarrow\left\{{}\begin{matrix}AB=BC=CD\\\widehat{A_1}=\widehat{B_1}=\widehat{C_1}=60^o\end{matrix}\right.\) (ĐN tam giác đều)
Vì AB = BD (gt), AC = CE (gt)
mà AB = BC = CD (cmt)
=> AB = AC = DB = BC = CE
Ta có: \(\widehat{B_1}+\widehat{B_2}=180^o\) (2 góc kề bù)
mà \(\widehat{B_1}=60^o\) (cmt)
=> \(\widehat{B_2}=120^o\)
CMTT, ta có: \(\widehat{C_2}=120^o\)
Xét \(\Delta ABD\) và \(\Delta ACE\) có:
AB = AC (cmt)
\(\widehat{B_2}=\widehat{C_2}\left(=120^o\right)\)
BD = CE (cmt)
=> \(\Delta ABD\) = \(\Delta ACE\) (c.g.c)
b) Ta có: \(\widehat{B_2}+\widehat{D}+\widehat{A_2}=180^o\) (t/c tam giác)
mà \(\widehat{B_2}=120^o\) (cmt)
=> \(\widehat{D}+\widehat{A_2}=60^o\)
mà \(\widehat{D}=\widehat{A_2}\) (\(\Delta ABD\) cân tại B do AB=BD)
=> \(\widehat{A_2}=30^o\)
CMTT, ta có: \(\widehat{A_3}=30^o\)
Lại có: \(\widehat{A_1}+\widehat{A_2}+\widehat{A_3}=\widehat{DAE}\)
=> \(\widehat{DAE}=30^o+60^o+30^o=150^o\)
c) Vì \(\Delta ABD\) = \(\Delta ACE\) (cmt)
=> AD = AE (2 cạnh tương ứng)
=> \(\Delta ADE\) cân tại A
