Giải thích các bước giải:
a.Ta có: $\widehat{KMA}=\widehat{BMN} $ (đối đỉnh)
$\widehat{MBN}=180^o-\widehat{MNB}-\widehat{BMN}=180^o-90^o-\widehat{BMN}=180^o-\widehat{KAM}-\widehat{KMA}=\widehat{MKA}$
Mà $\widehat{MBN}=\widehat{ABC},\widehat{NKC}=\widehat{MKA}$
$\to \widehat{ABC}=\widehat{NKC}$
b.Ta có:
$\widehat{AMN}=180^o-\widehat{BMN}=180^o-(90^o-\widehat{MBN})=180^o-(90^o-\widehat{ABC})=180^o-\widehat{ACB}$
$\to \widehat{AMN},\widehat{ACB}$ bù nhau