Trên tia đối của HB lấy D sao cho HD = HB
\(\rightarrow\) BC - DA
Chứng minh được:
BC = BD; \(\widehat{HBC}=\widehat{HBD}\)
\(\rightarrow\) BD = AD
\(\rightarrow\) \(\widehat{DBA}=\widehat{DAB}=\widehat{HBD}\) (Cùng phụ với \(\widehat{BCA}\))
\(\rightarrow\) \(\widehat{DBA}=\widehat{DAB}=\widehat{HBC}=\widehat{HBD}\)
Có: \(\widehat{CDB}=\widehat{DBA}+\widehat{DAB}=\widehat{HBC}+\widehat{HBD}\)
\(\rightarrow\widehat{CBD}=\widehat{CDB}\)
\(\rightarrow\widehat{CBD}=\widehat{CDB}=\widehat{BCD}=60^o\)
\(\rightarrow\widehat{CAB}=30^o\)
Vậy \(\widehat{BCA}=60^o\) và \(\widehat{CAB}=30^o\)