Vì \(\Delta\)ABC cân tại A \(\rightarrow AB=AC;\widehat{ABC}=\widehat{ACB}\)
Xét \(\Delta AHB\) và \(\Delta AKC\) có:
\(\widehat{A}\) chung
AH=AK (gt)
AB=AC
\(\rightarrow\Delta AHB=\Delta AKC\left(cgc\right)\)
\(\rightarrow\widehat{ABH}=\widehat{ACK}\) (2 góc tương ứng)
Ta có: \(\widehat{ABC}=\widehat{OBC}+\widehat{ABH}\); \(\widehat{ACB}=\widehat{OCB}+\widehat{ACK}\)
\(\widehat{ABC}+\widehat{ACB}\rightarrow\widehat{OBC}+\widehat{ABH}=\widehat{OCB}+\widehat{ACK}\)
mà \(\widehat{ABH}=\widehat{ACK}\)
\(\rightarrow\widehat{OBC}=\widehat{OCB}\)
\(\Rightarrow\Delta OBC\) cân tại O (đpcm)