Xét tứ giác $ABCD$:
$\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^o$
mà $\widehat{A}+\widehat{B}=220^o$
$→\widehat{C}+\widehat{D}=360^o-220^o=140^o$
$CO$ là đường phân giác $\widehat{C}$
$→\widehat{OCD}=\dfrac{1}{2}.\widehat{C}$
$DO$ là đường phân giác $\widehat{D}$
$→\widehat{ODC}=\dfrac{1}{2}.\widehat{D}$
mà $180^o-\widehat{OCD}+\widehat{ODC}=\widehat{COD}$
$→180^o-\dfrac{1}{2}.\widehat{C}+\dfrac{1}{2}.\widehat{D}=\widehat{COD}$
$→180^o-\dfrac{1}{2}.(\widehat{C}+\widehat{D})=\widehat{COD}$
$→180^o-\dfrac{1}{2}.140^o=\widehat{COD}$
$→180^o-70^o=\widehat{COD}$
$→110^o=\widehat{COD}$
Vậy $\widehat{COD}=110^o$
