a) Ta có : \(\widehat{ABC}=\widehat{ACB}\) (ΔABC cân tại A)
Lại có : \(\widehat{ACB}=\widehat{ECN}\) (đối đỉnh)
Suy ra : \(\widehat{ABC}=\widehat{ECN}\left(=\widehat{ACB}\right)\)
Xét \(\Delta MBD,\Delta NCE\) có :
\(\widehat{MBD}=\widehat{NCE}\) [do \(\widehat{ABC}=\widehat{ECN}\left(cmt\right)\)]
\(BD=CE\) (gt)
\(\widehat{MDB}=\widehat{NEC}\left(90^o\right)\)
=> \(\Delta MBD=\Delta NCE\left(c.g.c\right)\)
=> \(DM=NE\) (2 cạnh tương ứng)
b) Xét \(\Delta MID,\Delta NIE\) có :
\(\widehat{MDI}=\widehat{NEI}\left(=90^{^O}\right)\)
\(DM=NE\left(cmt\right)\)
\(\widehat{MID}=\widehat{NIE}\) (đối đỉnh)
=> \(\Delta MID=\Delta NIE\left(g.c.g\right)\)
=> \(IM=IN\) (2 cạnh tương ứng).
