Giải thích các bước giải:

a.Ta có $\Delta ABM,\Delta ACN$ đều

$\to\widehat{MAB}=60^o,\widehat{CAN}=60^o$

$\to\widehat{MAN}=\widehat{MAB}+\widehat{BAC}+\widehat{CAN}=180^o$

$\to A, M, N$ thẳng hàng

b.Ta có: $\Delta AMB,\Delta ANC$ đều

$\to AM=AB=BM, AN=AC=CN$

Xét $\Delta AMC,\Delta ABN$ có:

$AM=AB$

$\widehat{MAC}=\widehat{MAB}+\widehat{BAC}=120^o=\widehat{BAC}+\widehat{CAN}=\widehat{BAN}$

$AC=AN$

$\to\Delta AMC=\Delta ABN(c.g.c)$

$\to CM=BN$

c.Gọi $AB\cap CM=D$

Từ câu b $\to\widehat{AMC}=\widehat{ABN}$

$\to\widehat{AMD}=\widehat{DBO}$

Mà $\widehat{ADM}=\widehat{BDO}$(đối đỉnh)

$\to\widehat{DOB}=180^o-\widehat{ODB}-\widehat{OBD}=180^o-\widehat{ADM}-\widehat{DMA}=\widehat{MAD}=60^o$

$\to\widehat{BOC}=180^o-\widehat{DOB}=120^o$