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Giải thích các bước giải:

a.Vì $O\in$ trung trực $AB, AC\to OD\perp AB, OE\perp AC$

Xét $\Delta ODA,\Delta ODB$ có:

Chung $OD$

$\widehat{ODA}=\widehat{ODB}(=90^o)$

$DA=DB$

$\to\Delta ODA=\Delta ODB(c.g.c)$

b.Vì $N\in$ trung trực $AC\to NA=NC$

         $O\in$ trung trực $AC\to OA=OC$

Xét $\Delta NOA,\Delta NOC$ có:
Chung $ON$

$OA=OC$
$NA=NC$

$\to\Delta NOA=\Delta NOC(c.c.c)$

c.Vì $N, O\in$ trung trực $AB\to MA=MB, OA=OB$

Xét $\Delta AOM,\Delta BOM$ có:
Chung $OM$

$OA=OB$

$MA=MB$

$\to\Delta AOM=\Delta BOM(c.c.c)$

$\to \widehat{MAO}=\widehat{OBM}=\widehat{OBC}$

Do $OA=OB, OA=OC\to OB=OC\to\Delta OBC$ cân tại $O$

$\to \widehat{OBC}=\widehat{OCB}$

$\to\widehat{OAM}=\widehat{OAB}$

$\to AO$ là phân giác $\widehat{MAN}$

$\to AO$ là phân giác $\widehat{MAN}$

d.Ta có:

$\widehat{OBC}+\widehat{OCB}$

$=(\widehat{ABC}-\widehat{OBA})+(\widehat{ACB}-\widehat{OCB})$

$=\widehat{ABC}+\widehat{ACB}-(\widehat{OBA}+\widehat{OCA})$

$=180^o-\widehat{BAC}-(\widehat{OAB}+\widehat{OAC})$

$=180^o-\widehat{BAC}-\widehat{BAC}$

$=180^o-2\widehat{BAC}$

$=60^o$

Từ câu c 

$\to \widehat{MAN}=2\widehat{OAN}=2\widehat{OBC}=\widehat{OBC}+\widehat{OCB}=60^o$