Giải thích các bước giải:
Ta có : $\widehat{CBA}=60^o+20^o=80^o$
$\widehat{BCA}=50^o+30^o=80^o\to\widehat{BAC}=20^o$
Vẽ $BG$ sao cho $\widehat{CBG}=20^o, G\in AC$
Vì $\widehat{ACG}=80^o\to\widehat{BGC}=80^o, \widehat{GBE}=\widehat{CBA}-\widehat{CBG}-\widehat{EBF}=40^o$
$\to \widehat{BGE}=\widehat{GBC}+\widehat{BCG}=100^o$
$\to \widehat{BEG}=40^o\to BG=GE$
Xét $\Delta BCF, \widehat{CBF}=80^o,\widehat{BCF}=50^o\to \widehat{BFC}=50^o$
$\to BF=BC\to BF=BG(BC=BG)$
Mà $\widehat{GBF}=\widehat{GBE}+\widehat{EBF}=60^o\to\Delta BFG$ đều
$\to\widehat{BGF}=60^o\to \widehat{FGE}=180^o-\widehat{BGC}-\widehat{BGF}=40^o$
Mà $GF=BG=GE\to\Delta GEF$ cân tại G
$\to\widehat{GEF}=\dfrac12(180^o-\widehat{FGE})=70^o$
$\to x+\widehat{BEG}=\widehat{FEG}=70^o\to x=30^o$
