Giải thích các bước giải:
Ta có:
$\widehat{BDC}=\widehat{DBA}+\widehat{DAB}$
$\to \widehat{BDC}=\dfrac12\widehat{ABC}+\widehat{BAC}$ vì $BD$ là phân giác $\widehat{ABC}$
$\to \widehat{BDC}=\dfrac12\widehat{ABC}+60^o$
$\to \widehat{BDC}=\dfrac12(180^o-\widehat{BAC}-\widehat{ACB})+60^o$
$\to \widehat{BDC}=\dfrac12(180^o-60^o-\widehat{ACB})+60^o$
$\to \widehat{BDC}=\dfrac12(120^o-\widehat{ACB})+60^o$
$\to \widehat{BDC}=60^o-\dfrac12\widehat{ACB}+60^o$
$\to \widehat{BDC}=120^o-\dfrac12\widehat{ACB}$
$\to \widehat{BDC}=120^o-\widehat{ACE}$ vì $CE$ là phân giác $\widehat{ACB}$
$\to \widehat{BDC}=180^o-60^o-\widehat{ACE}$
$\to \widehat{BDC}=180^o-\widehat{EAC}-\widehat{ACE}$
$\to \widehat{BDC}=\widehat{AEC}$
$\to đpcm$
