Giải thích các bước giải:
Vì $\widehat{BAC}=45^o\rightarrow\widehat{BOC}=2\widehat{BAC}=90^o$
Gọi $CE\cap BD=H, CH\cap AB=F, BD\cap AC=G$
$\rightarrow \widehat{EOB}=2\widehat{ECB},\widehat{COD}=2\widehat{DBC}$
$\begin{split}\to\widehat{EOB}+\widehat{COD}&=2\widehat{ECB}+2\widehat{DBC}\\&=2.(\widehat{HCB}+\widehat{HBC})\\&=2\widehat{GHC}\\&=2\widehat{BAC}\\&=90^o\end{split}$
$\rightarrow\widehat{EOB}+\widehat{COD}+\widehat{BOC}=180^o$
$\rightarrow\widehat{EOD}=180^o$
$\rightarrow E,O,D$ thẳng hàng