Chứng minh:

a)

$\widehat{MAC}=\widehat{MAB}+\widehat{BAC}=90∘+\widehat{BAC}$;

$\widehat{NAB}=\widehat{NAC}+\widehat{BAC}=90∘+\widehat{BAC}$;

=> $\widehat{MAC}=\widehat{NAB}$

Xét tam giác AMC và ABN ta có:

AM = AB 

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

AC = AN

=> ΔMAC=ΔNAB (c.g.c)

b)

ΔMAC=ΔNAB => $\widehat{ACM}=\widehat{ANB}$

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

$\widehat{KIC}+\widehat{KCI}=\widehat{AIN}+\widehat{ANI}=90^{\circ}$

=>$\widehat{IKC}=90^{\circ}$

Vậy BN⊥MC