a) ABCD là hình vuông \(\Rightarrow\)AD = AB = BC = CD; \(\widehat{A}\)= \(\widehat{B}\)= \(\widehat{C}\)= \(\widehat{D}\)= 900
Xét \(\Delta\)ADF và \(\Delta\)CDE có:
\(\widehat{FAD}\)= \(\widehat{ECD}\) ( = 150 )
AD = CD
\(\widehat{FDA}\)= \(\widehat{EDC}\)
suy ra \(\Delta\)ADF = \(\Delta\)CDE (g.c.g)
\(\Rightarrow\)DF = DE
\(\Rightarrow\)\(\Delta\)DEF cân tại D
Ta lại có: \(\widehat{A}\)= \(\widehat{ADF}\)+ \(\widehat{CDE}\)+ \(\widehat{EDF}\)
\(\Rightarrow\)900 = 150 + 150 + \(\widehat{EDF}\)
\(\Rightarrow\)\(\widehat{EDF}\)= 900 - 150 - 150 = 600
suy ra \(\Delta\)DEF là tam giác đều