Giải thích các bước giải:

a.Xét $\Delta OAD,\Delta OBC$ có:

$OA=OC$

Chung $\hat O$

$OD=BO$

$\to \Delta OAD=\Delta OCB(c.g.c)$

$\to AD=BC$

b.Từ a $\to \widehat{OCB}=\widehat{OAD}, \widehat{ODA}=\widehat{OBC}$

$\to \widehat{ICD}=180^o-\widehat{OCB}=180^o-\widehat{OAD}=\widehat{IAB}$

      $\widehat{IDC}=\widehat{ADO}=\widehat{OBC}=\widehat{IBA}$

Xét $\Delta AIB,\Delta CID$ có:

$\widehat{IBA}=\widehat{IDC}$

$AB=OB-OA=OD-OC=CD$

$\widehat{IAB}=\widehat{ICD}$

$\to \Delta AIB=\Delta CID(g.c.g)$

c.Từ b $\to IA=IC$

Xét $\Delta OIA,\Delta OIC$ có:
Chung $OI$

$OA=OC$

$IA=IC$

$\to \Delta OIA=\Delta OIC(c.c.c)$

$\to \widehat{IOA}=\widehat{IOC}$

$\to OI$ là phân giác $\widehat{xOy}$

d.Ta có: $OA=OC, OB=OD$

$\to \Delta OAC,\Delta OBD$ cân tại $O$

$\to \widehat{OAC}=90^o-\dfrac12\hat O=\widehat{OBD}$

$\to AC//BD$