Giải thích các bước giải:

a.Vì $DF//BC\to\widehat{AFD}=\widehat{ABC}, \widehat{ADF}=\widehat{ACB}$

Mà $\widehat{ABC}=\widehat{ACB}\to \widehat{AFD}=\widehat{ADF}\to AF=AD$

Lại có $AB=AC\to\Delta AFC=\Delta ADB(c.g.c)$

b.Từ câu a $\to \widehat{ABD}=\widehat{ACF}$

$\to \widehat{OBC}=\widehat{ABC}-\widehat{ABD}=\widehat{ACB}-\widehat{ACF}=\widehat{OCB}$

$\to\widehat{OCB}=\widehat{OBC}=60^o\to\Delta OBC$ đều

Do $DF//BC\to\widehat{ODF}=\widehat{OBC}=60^o,\widehat{DFO}=\widehat{OCB}=60^o\to\Delta ODF$ đều

c.Ta có $\Delta ABC$ cân tại A
$\to \widehat{ABC}=\widehat{ACB}=90^o-\dfrac12\widehat{BAC}=80^o$ 

Ta có $\widehat{ABF}=\widehat{ABC}-\widehat{CBD}=80^o-60^o=20^o$

Vì $\widehat{BCE}=50^o\to \widehat{BEC}=180^o-\widehat{EBC}-\widehat{ECB}=180^o-80^o-50^o=50^o$

$\to\widehat{BEC}=\widehat{BCE}\to\Delta BCE$ cân tại B

$\to BE=BC\to BE=BO(\Delta OBC\quad đều )$

$\to \widehat{BEO}=\widehat{BOE}=90^o-\dfrac12\widehat{EBO}=90^o-\dfrac12.20^o=80^o$

$\to\widehat{FEO}=180^o-\widehat{BEO}=100^o$

      $\widehat{FOE}=180^o-\widehat{EOB}-\widehat{BOC}=180^o-80^o-60^o=40^o$

$\to \widehat{EFO}=180^o-\widehat{FEO}-\widehat{FOE}=180^o-100^o-40^o=40^o$

Ta có $\widehat{EFO}=\widehat{EOF}\to \Delta EFO$ cân tại E

$\to EO=EF$

Mà $\Delta OFD $ đều $\to OD=DF\to\Delta EFD=\Delta EOD(c.c.c)$

$\to \widehat{OED}=\widehat{FED}=\dfrac12\widehat{OEF}=50^o$

      $\widehat{EDO}=\widehat{FDE}=\dfrac12\widehat{FDO}=30^o$

      $\widehat{EOD}=\widehat{EOF}+\widehat{FOD}=40^o+60^o=100^o$

d.Từ câu c $\to \Delta EFD=\Delta EOD\to đpcm$