Question

Answer 1

1) Xét ΔDAC và ΔBAE có
AD = AB
AC = AE
\(\widehat{DAE}=\widehat{BAC}\)
=> ΔDAC = ΔBAE (c.g.c)
=> DC = BE
2) Ta có \(\widehat{AEB}=\widehat{EAC}=90^o\)
mà \(\widehat{EAC}=\widehat{BAC}\)
\(\widehat{ACD}=\widehat{AEB}=>\widehat{BAC}+\widehat{AEB}=90^o\)
\(=>\widehat{EOC}=90^o=>BE\perp DC\)