Giải thích các bước giải:
Xét tứ giác $ABCD$ nội tiếp $(O)$
$\to \widehat{ABC}+\widehat{BCD}+\widehat{CDA}+\widehat{DAB}=360^o$
$\to \widehat{ABC}+(\widehat{BCO}+\widehat{OCD})+\widehat{CDA}+(\widehat{DAO}+\widehat{OAB})=360^o$
$\to \widehat{ABC}+\widehat{CDA}+(\widehat{BCO}+\widehat{OCD})+(\widehat{DAO}+\widehat{OAB})=360^o$
$\to \widehat{ABC}+\widehat{CDA}+\widehat{BCO}+\widehat{OCD}+\widehat{DAO}+\widehat{OAB}=360^o$
$\to \widehat{ABC}+\widehat{CDA}+\widehat{CBO}+\widehat{ODC}+\widehat{ODA}+\widehat{OBA}=360^o$
$\to \widehat{ABC}+\widehat{CDA}+(\widehat{CBO}+\widehat{OBA})+(\widehat{ODC}+\widehat{ODA})=360^o$
$\to \widehat{ABC}+\widehat{CDA}+\widehat{ABC}+\widehat{ADC}=360^o$
$\to 2(\widehat{ABC}+\widehat{CDA})=360^o$
$\to \widehat{ABC}+\widehat{CDA}=180^o$
$\to $Trong một tứ giác nội tiếp thì tổng hai góc đối diện bằng $180^o$
