a) Ta có: [tex]\widehat{ACB}=\widehat{KCN}=90^o\Rightarrow \widehat{ACK}=\widehat{BCN}[/tex]
Lại có: [tex]\widehat{CAK}=\widehat{CBN}=\frac{1}{2}sđBC[/tex] nên [tex]\Delta CAK\sim \Delta CBN[/tex]
b) Ta có: [tex]\Delta CAK\sim \Delta CBN\Rightarrow \frac{CK}{CN}=\frac{CA}{CB}[/tex]
Chứng minh tương tự ta có: [tex]\Delta CAM\sim \Delta CBK\rightarrow \frac{CA}{CB}=\frac{CM}{CK}\Rightarrow \frac{CM}{CK}=\frac{CK}{CN}\Rightarrow \Delta CMK\sim \Delta CKN\Rightarrow \widehat{CMK}=\widehat{CKN}\Rightarrow \widehat{CMK}+\widehat{CKM}=\widehat{CKM}+\widehat{CKN}\Rightarrow \widehat{MKN}=90^o[/tex]