a) Ta có: $\widehat{MAC}=\widehat{MAB}+\widehat{BAC}=90^{\circ}+\widehat{BAC}; \widehat{NAB}=\widehat{NAC}+\widehat{BAC}=90^{\circ}+\widehat{BAC}$

Suy ra $\widehat{MAC}=\widehat{NAB}$

Xét tam giác AMC và ABN ta có:

AM = AB 

 $\widehat{MAC}=\widehat{NAB}$

AC = AN

Suy ra $\Delta MAC =\Delta NAB$ (c.g.c)

b) Do $\Delta MAC =\Delta NAB$ nên $\widehat{ACM}=\widehat{ANB}$.

Mặt khác $\widehat{AIN}=\widehat{KIC}$ (đối đỉnh)

Xát tam giác KIC, có $\widehat{KIC}+\widehat{KCI}=\widehat{AIN}+\widehat{ANI}=90^{\circ}$.

Suy ra $\widehat{IKC}=90^{\circ}$

Vậy $BN\perp MC$