Giải thích các bước giải:

a.Xét $\Delta OAC,\Delta OBC$ có:

Chung $OC$

$\widehat{COA}=\widehat{COB}$

$OA=OB$

$\to \Delta OAC=\Delta OBC(c.g.c)$

$\to \widehat{OCA}=\widehat{OCB}, CA=CB$

$\to C$ là trung điểm $AB$

Vì $\widehat{OCA}+\widehat{OCB}=180^o$

$\to \widehat{OCA}=\widehat{OCB}=90^o$

$\to OC\perp AB$

b.Xét $\Delta COB,\Delta CMA$ có:

$CB=CA$

$\widehat{BCO}=\widehat{ACM}$

$CO=CM$

$\to \Delta COB=\Delta CMA(c.g.c)$
$\to \widehat{COB}=\widehat{CMA}$

$\to AM//OB$

Xét $\Delta CBM,\Delta COA$ có:

$CB=CA$

$\widehat{BCM}=\widehat{OCA}$

$CM=CO$

$\to \Delta CBM=\Delta CAO(c.g.c)$

$\to \widehat{CBM}=\widehat{CAO}$

$\to BM//OA$

c.Từ câu b $\to MA=OB, MB=OA$

Vì $OA=OB\to MB=MA$

Ta có: $M\in$ phân giác $\widehat{xOy}, MK\perp Ox, MI\perp Oy$

$\to MK=MI$

$\to AK^2=MA^2-MK^2=MB^2-MI^2=BI^2$

$\to AK=BI$

d.Xét $\Delta OAI,\Delta OBK$ có:

$OA=OB$

Chung $\hat O$

$OI=OB+BI=OA+AK=OK$

$\to \Delta OAI=\Delta OBK(c.g.c)$

$\to \widehat{OAI}=\widehat{OBK}\to \widehat{IBN}=180^o-\widehat{OBK}=180^o-\widehat{OAI}=\widehat{NAK}$

      $\widehat{OIA}=\widehat{OKB}\to \widehat{BIN}=\widehat{NKA}$

Xét $\Delta NBI,\Delta NAK$ có:

$\widehat{NBI}=\widehat{NAK}$

$BI=AK$

$\widehat{NIB}=\widehat{NKA}$

$\to \Delta NBI=\Delta NAK(g.c.g)$

$\to NI=NK$

Ta có: $MI=MK, NI=NK, OI=OK$

$\to O, N, M\in$ trung trực $IK$

$\to O, N, M$ thẳng hàng