Giải thích các bước giải:
Ta có: $BO,EO$ là phân giác $\widehat{ABC},\widehat{AED}$
$\to \widehat{CBO}=\widehat{OBA}=\dfrac12\widehat{ABC}$
$\widehat{AEO}=\widehat{OED}=\dfrac12\widehat{AED}$
$\to \widehat{OBE}+\widehat{OEB}=(\widehat{OBA}+\widehat{ABE}) +(\widehat{OEA}+\widehat{AEB})$
$\to \widehat{OBE}+\widehat{OEB}=\widehat{OBA}+\widehat{ABE}+\widehat{OEA}+\widehat{AEB}$
$\to \widehat{OBE}+\widehat{OEB}=(\widehat{OBA}+\widehat{OEA})+(\widehat{ABE}+\widehat{AEB})$
$\to 180^o-\widehat{BOE}=(\dfrac12\widehat{ABC}+\dfrac12\widehat{AED})+(180^o-\widehat{BAE})$
$\to 180^o-\widehat{BOE}=\dfrac12(\widehat{ABC}+\widehat{AED})+180^o-\widehat{BAE}$
$\to -\widehat{BOE}=\dfrac12(\widehat{ABC}+\widehat{AED})-\widehat{BAE}$
$\to \widehat{BOE}=\widehat{BAE}-\dfrac12(\widehat{ABC}+\widehat{AED})$
$\to \widehat{BOE}=\dfrac12(2\widehat{BAE}-\widehat{ABC}-\widehat{AED})$
$\to \widehat{BOE}=\dfrac12((\widehat{BAE}-\widehat{ABC})+(\widehat{BAE}-\widehat{AED}))$
$\to \widehat{BOE}=\dfrac12((\widehat{ABC}+\widehat{ACB}-\widehat{ABC})+(\widehat{ADE}+\widehat{AED}-\widehat{AED}))$
$\to \widehat{BOE}=\dfrac12(\widehat{ACB}+\widehat{ADE})$
$\to \widehat{BOE}=\dfrac12(\widehat{ECB}+\widehat{EDB})$
