Giải thích các bước giải:
Trước hết ta chứng minh tổng các góc trong một tứ giác bằng $360^o$
Thật vậy xét tứ giác $ABCD$
$\to \widehat{ABC}+\widehat{BCD}+\widehat{CDA}+\widehat{DAB}=(\widehat{ABD}+\widehat{DBC})+\widehat{BCD}+(\widehat{CDB}+\widehat{BDA})+\widehat{DAB}$
$\to \widehat{ABC}+\widehat{BCD}+\widehat{CDA}+\widehat{DAB}=\widehat{ABD}+\widehat{DBC}+\widehat{BCD}+\widehat{CDB}+\widehat{BDA}+\widehat{DAB}$
$\to \widehat{ABC}+\widehat{BCD}+\widehat{CDA}+\widehat{DAB}=(\widehat{ABD}+\widehat{BDA}+\widehat{DAB})+(\widehat{BDC}+\widehat{BCD}+\widehat{DBC})$
$\to \widehat{ABC}+\widehat{BCD}+\widehat{CDA}+\widehat{DAB}=180^o+180^o$
$\to \widehat{ABC}+\widehat{BCD}+\widehat{CDA}+\widehat{DAB}=360^o$
a.Áp dụng tính chất trên có:
$\hat{I_1}=180^o-\hat{I_2}$
$\to \hat{I_1}=360^o-90^o-90^o-\hat{I_2}$
$\to\hat{I_1}=\hat O$
b.Từ câu a ta có:
$\hat{I_1}=\hat O$
$\to 180^o-\hat{I_2}=\hat O$
$\to\hat{I_2}+\hat O+180^o$
