Giải thích các bước giải:

a.Xét $\Delta OAM,\Delta OBM$ có:

Chung $OM$

$MA=MB$
$OA=OB$

$\to \Delta OMA=\Delta OMB(c.c.c)$

b.Từ câu a $\to \widehat{MOA}=\widehat{MOB}$

$\to \widehat{HOA}=\widehat{HOB}$

Xét $\Delta OAH,\Delta OBH$ có:

Chung $OH$

$\widehat{HOA}=\widehat{HOB}$

$OA=OB$

$\to\Delta HOA=\Delta HOB(c.g.c)$

c.Ta có: $OA=OB\to \Delta OAB$ cân tại $O$

               $AB//KE$

$\to \widehat{OKE}=\widehat{OBA}=\widehat{OAB}=\widehat{OEK}$

$\to\Delta OEK$ cân tại $O\to OE=OK$

$\to BK=OK-OB=OE-OA=EK$

Xét $\Delta AOK,\Delta BOE$ có:

$OA=OB$

Chung $\hat O$

$OK=OE$

$\to\Delta OAK=\Delta OBE(c.g.c)$

$\to \widehat{OKA}=\widehat{OEB}, \widehat{OAK}=\widehat{OBE}$

$\to \widehat{NKB}=\widehat{NEA},\widehat{NBK}=180^o-\widehat{OBE}=180^o-\widehat{OAK}=\widehat{NAE}$

Xét $\Delta NBK,\Delta NAE$ có:

$\widehat{NKB}=\widehat{NEA}$

$BK=AE$

$\widehat{NBK}=\widehat{NAE}$

$\to \Delta NBK=\Delta NAE(g.c.g)$

$\to NB=NA$

Mà $MB=MA, OB=OA$

$\to O, M, N\in$ trung trực $AB$

$\to O,M,N$ thẳng hàng