Xét \(\Delta ANB\) và\(\Delta BMC\) có:
\(\widehat{NAB}=\widehat{MBC}\left(=90^0\right)\)
\(\widehat{ABN}=\widehat{BCM}\) (cùng phụ với \(\widehat{KBC}\))
\(\Rightarrow\Delta ANB\infty\Delta BMC\)
\(\Rightarrow\frac{AN}{BM}=\frac{AB}{BC}=\frac{1}{2}\)
\(\Rightarrow AN=\frac{BM}{2}=\frac{a}{4}=\frac{AD}{8}\)
\(\Rightarrow k=4\)