Giải:
Xét \(\Delta DAC,\Delta BAE\) có:
\(AD=AB\) ( t/g ABD đều )
\(\widehat{DAC}=\widehat{BAE}\left(=60^o+\widehat{BAC}\right)\)
AC = AE ( t/g ACE đều )
\(\Rightarrow\Delta DAC=\Delta BAE\left(c-g-c\right)\)
\(\Rightarrow\widehat{ACD}=\widehat{AEB}\) ( góc t/ứng )
Ta có: \(\widehat{AEC}+\widehat{ACE}=120^o\) ( do \(\widehat{AEC}=\widehat{ACE}=60^o\) )
\(\Rightarrow\widehat{AEB}+\widehat{BEC}+\widehat{ACE}=120^o\)
\(\Rightarrow\widehat{ACD}+\widehat{ACE}+\widehat{MEC}=120^o\)
\(\Rightarrow\widehat{MCE}+\widehat{MEC}=120^o\)
Mà \(\widehat{CME}+\widehat{MCE}+\widehat{MEC}=180^o\)
\(\Rightarrow\widehat{CME}=60^o\)
\(\widehat{BMC}+\widehat{CME}=180^o\) ( kề bù )
\(\Rightarrow\widehat{BMC}=120^o\) hay \(\widehat{CMB}=120^o\)
Vậy...
