c) Ta có :
\(\widehat{ABE}=\widehat{HBD}\) (BE là phân giác \(\widehat{ABC}\))
mà \(\widehat{BAE}=\widehat{BHD}=90^o\)
\(\Rightarrow\Delta ABE\sim\Delta HBD\left(g.g\right)\)
\(\Rightarrow\dfrac{AB}{AE}=\dfrac{HB}{HD}\)
\(\Rightarrow\dfrac{AB}{2AM}=\dfrac{HB}{2DN}\) (\(M;N\) là trung điểm \(AE;DH\))
\(\Rightarrow\dfrac{AB}{AM}=\dfrac{HB}{DN}\)
mà \(\widehat{BAM}=\widehat{BHN}=90^o\)
\(\Rightarrow\Delta ABM\sim\Delta HBN\left(g.g\right)\)
\(\Rightarrow\widehat{ABM}=\widehat{NBH}\)
mà \(\widehat{ABM}+\widehat{MBE}=\widehat{ABE}\)
\(\widehat{NBH}+\widehat{NBD}=\widehat{CBE}\)
\(\widehat{CBE}=\widehat{ABE}\left(BE.là.phân.giác\right)\)
\(\Rightarrow\widehat{MBE}=\widehat{DBN}\)
\(\Rightarrow BE\) là phân giác \(\widehat{MBN}\left(đpcm\right)\)
