Đáp án:

Xét $\triangle OAC$ có: $OA=OC$ (gt)

$\Rightarrow \triangle OAC$ cân tại $O$

$\Rightarrow \widehat{OAC}=\widehat{OCA}$

mà $\widehat{OAC}+\widehat{BAC}=180^0$

$\widehat{OCA}+\widehat{ACD}=180^0$

$\Rightarrow \widehat{BAC}=\widehat{ACD}$

Xét $\triangle ABC$ và $\triangle CDA$ có:

$AC$ chung

$\widehat{BAC}=\widehat{ACD}$ (cmt)

$AB=CD$ (gt)

$\Rightarrow \triangle ABC=\triangle CDA$ (c.g.c)

$\Rightarrow \widehat{ABC}=\widehat{ADC}$ (hai góc tương ứng)

Ta có: $OB=OA+AB$

$OD=OC+CD$

mà $OA=AB=OC=CD$ (gt)

$\Rightarrow OB=OD$

$\Rightarrow \triangle OBD$ cân tại $O$

$\Rightarrow \widehat{OBD}=\widehat{ODB}$

mà $\widehat{ODB}=\widehat{ABC}+\widehat{KBD}$

$\widehat{ODB}=\widehat{ADC}+\widehat{KDB}$

$\widehat{ABC}=\widehat{ADC}$ (cmt)

$\Rightarrow \widehat{KBD}=\widehat{KDB}$ 

$\Rightarrow \triangle KBD $ cân tại $K$

$\Rightarrow KB=KD$ 

Xét $\triangle OKB$ và $\triangle OKD$ có

$OK$ chung

$KB=KB$ (cmt)

$OB=OD$ (cmt)

$\Rightarrow \triangle OKB=\triangle OKD$ (c.c.c)

$\Rightarrow \widehat{BOK}=\widehat{DOK}$

$\Rightarrow OK$ là phân giác góc $\widehat{xOy}$