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 $BN\perp MC$