c, Ta có DN=DB (cmb) ⇒ ΔDNB cân tại D ⇒\(\widehat{DNB}=\widehat{DBN}\)
Ta có DN=DE (=DB) ⇒ ΔDNE cân tại D ⇒\(\widehat{DNE}=\widehat{DEN}\)
Ta có:\(\widehat{DNB}+\widehat{DNE}=\widehat{DBN}+\widehat{DEN}\)
Xét ΔEBN có:\(\widehat{EBN}+\widehat{ENB}+\widehat{BEN}=180^o\)
\(\Rightarrow\widehat{DBN}+\widehat{DNB}+\widehat{DNE}+\widehat{DEN}=180^o\\ \Rightarrow\left(\widehat{DBN}+\widehat{DEN}\right)+\left(\widehat{DNB}+\widehat{DNE}\right)=180^o\\ \Rightarrow2\widehat{BNE}=180^o\\ \Rightarrow\widehat{BNE}=90^o\\ \Rightarrow BN\perp Ex\)