Giải thích các bước giải:
1.Ta có :
$AD\perp BC, CF\perp AB\to\widehat{HFB}+\widehat{HDB}=90^o+90^o=180^o$
$\to BFHD$ nội tiếp
$\to \widehat{AHC}=\widehat{FHD}=180^o-\widehat{ABC}$
2.Vì $M,N$ đối xứng qua AC
$\to\widehat{ANC}=\widehat{AMC}=\widehat{ABC}$
$\to \widehat{AHC}+\widehat{ANC}=180^o-\widehat{ABC}+\widehat{ABC}=180^o$
$\to AHCN$ nội tiếp
3.Ta có :
$\widehat{IAJ}=\widehat{MAC}=\widehat{CAN}=\widehat{CHN}=\widehat{JHI}$
$\to AHIJ$ nội tiếp
$\to \widehat{ẠJI}=180^o-\widehat{AHI}=180^o-\widehat{AHC}=\widehat{ANC}$
4.Gọi $BH\cap AC=E$
Vì $CF\perp AB,AD\perp BC\to H$ là trực tâm $\Delta ABC$
$\to BE\perp AC\to AEHF$ nội tiếp
$\to \widehat{EFH}=\widehat{HAE}=\widehat{JIC}$ vì IJAH nội tiếp
$\to EF//IJ$
Lại có :
$\widehat{OAC}=\widehat{OCA}=90^o-\dfrac12\widehat{AOC}=90^o-\widehat{ABC}=\widehat{FAH}=\widehat{HEF}$
$\to OA\perp EF\to OA\perp IJ$