Giải thích các bước giải:

Xét $\Delta CDN,\Delta CMB$ có:

$CD=BM(=AB)$

$\widehat{NDC}=60^o+\widehat{ADC}=60^o+\widehat{ABC}=\widehat{MBC}$

$DN=BC(=AD)$

$\to \Delta CDN=\Delta MBC(c.g.c)$

$\to CN=CM, \widehat{CND}=\widehat{MCB}$

$\to \widehat{MCN}=\widehat{BCD}-(\widehat{MCB}+\widehat{NCD})$

$\to \widehat{MCN}=\widehat{BCD}-(\widehat{DNC}+\widehat{NCD})$

$\to \widehat{MCN}=\widehat{BCD}-(180^o-\widehat{NDC})$

$\to \widehat{MCN}=\widehat{BCD}-(180^o-\widehat{ADC}-60^o)$

$\to \widehat{MCN}=\widehat{BCD}+\widehat{ADC}-120^o$

$\to \widehat{MCN}=180^o-120^o=60^o$

$\to \Delta CMN$ đều