\(\widehat{ABH}=\widehat{BAC}+\widehat{BDA}=\widehat{BAC}+90^o\) (tính chất góc ngoài)
\(\widehat{ACK}=\widehat{BAC}+\widehat{CEA}=\widehat{BAC}+90^o\) (tính chất góc ngoài)
\(\Rightarrow\widehat{ABH}=\widehat{ACK}\)
Theo đề bài: \(AB=CK,BH=AC\)
\(\Delta ABH=\Delta KAC\) (c.g.c)
Vậy: \(AH=AK\)