Giải thích các bước giải:
a.Ta có $AH\perp BC, BK\perp AC$
$\to\Delta AHC,\Delta BKC$ vuông tại $H,K$
$\to \widehat{HAC}=90^o-\widehat{HCA}=90^o-\widehat{BCK}=\widehat{KBC}$
b.Xét $\Delta ACM,\Delta CBN$ có:
$AC=BN$
$\widehat{MAC}=180^o-\widehat{HAC}=180^o-\widehat{KBC}=\widehat{NBC}$
$AM=BC$
$\to\Delta CBN=\Delta MAC(c.g.c)$
c.Từ câu b
$\to\widehat{BCN}=\widehat{AMC}=\widehat{HMC}$
$\to \widehat{NCM}=\widehat{NCB}+\widehat{BCM}=\widehat{HMC}+\widehat{HCM}=90^o$ vì $AH\perp BC$
$\to CM\perp CN$