Giải thích các bước giải:
a.Ta có $\widehat{MAN}=\widehat{MAB}+\widehat{BAC}+\widehat{CAN}=60^o+60^o+60^o=180^o$
$\to M,A,N$ thẳng hàng
b.Xét $\Delta AMC,\Delta ABN$ có:
$AM=AB$ vì $\Delta ABM$ đều
$\widehat{MAC}=\widehat{MAB}+\widehat{BAC}=\widehat{NAC}+\widehat{ABC}=\widehat{BAN}$
$AC=AN$ vì $\Delta ACN$ đều
$\to \Delta AMC=\Delta ABN(c.g.c)$
$\to CM=BN$
c.Từ câu b
$\to \widehat{ABN}=\widehat{AMC}$
Gọi $AB\cap CM=D$
$\to\widehat{OBD}=\widehat{DMA}$
$\to \widehat{DOB}=180^o-\widehat{BDO}-\widehat{DBO}=180^o-\widehat{MDA}-\widehat{DMA}=\widehat{MAD}=60^o$
$\to\widehat{BOC}=180^o-\widehat{DOB}=120^o$