Giải thích các bước giải:
a.Xét $\Delta OHA,\Delta OHB$ có:
$\widehat{BHO}=\widehat{OHA}=90^o$ vì $AB\perp Ot$ tại $H$
Chung $OH$
$\widehat{BOH}=\widehat{HOA}$ vì $Ot$ là phân giác $\widehat{xOy}$
$\to\Delta OHA=\Delta OHB(g.c.g)$
$\to OA=OB$
b.Xét $\Delta ODC,\Delta OEC$ có:
$OD=OE$
$\widehat{DOC}=\widehat{COE}$ vì $Ot$ là phân giác $\widehat{xOy}$
Chung $OC$
$\to\Delta ODC=\Delta OEC(c.g.c)$
$\to CD=CE,\widehat{CDO}=\widehat{CEA}$
$\to\widehat{BDC}=180^o-\widehat{ODC}=180^o-\widehat{CEO}=\widehat{CEA}$
Ta có $OA=OB,OD=OE\to BD=OB-OD=OA-OE=AE$
Xét $\Delta BCD,\Delta ACE$ có:
$CD=CE$
$\widehat{BDC}=\widehat{CEA}$
$BD=AE$
$\to\Delta BCD=\Delta ACE(c.g.c)$
$\to\widehat{BCD}=\widehat{ACE}$
$\to \widehat{BCE}=\widehat{BCA}+\widehat{ACE}=\widehat{BCA}+\widehat{BCD}=\widehat{ACD}=180^o$
$\to B,C,E$ thẳng hàng