Giải thích các bước giải:
a.Ta có :
$\begin{cases}AD=AB\\\widehat{DAC}=\widehat{BAE}(=\widehat{BAC}+60^o)\\ AC=AE\end{cases}$
$\rightarrow\Delta ADC=\Delta ABE(c.g.c)$
b.Từ câu a
$\rightarrow\widehat{ACD}=\widehat{AEB} $
Gọi $BE\cap AC=F\rightarrow\widehat{CAE}=\widehat{EMC}(\widehat{AFE}=\widehat{MFC},\widehat{AEF}=\widehat{FCM})$
$\rightarrow\widehat{FMC}=60^o$
$\rightarrow\widehat{BMC}=180^o-\widehat{FMC}=120^o$