Ta có:\(\widehat{BAK}+\widehat{KAC}=90^o\)
Xét ΔKAC ta có:
\(\widehat{KAC}+\widehat{KCA}+\widehat{AKC}=180^o\\ \Rightarrow90^o+\widehat{KAC}+\widehat{KCA}=180^o\\ \Rightarrow\widehat{KAC}+\widehat{KCA}=90^o\)
\(\Rightarrow\widehat{BAK}=\widehat{KCA}\)