Đáp án:

a)

Ta có:

$\widehat{DAC}=\widehat{BAC}+\widehat{DAB}$

$\widehat{BAE}=\widehat{BAC}+\widehat{CAE}$

mả $\widehat{DAB}=\widehat{CAE}=60^0$

$\Rightarrow \widehat{DAC}=\widehat{BAE}$

Xét $\triangle DAC$ và $\triangle BAE$ có:

$AD=AB$ (do $\triangle ABD$ đều-gt)

$AC=AE$ (do $\triangle ACE$ đều -gt)

$\widehat{DAC}=\widehat{BAE}$ (cmt)

$\Rightarrow \triangle DAC=\triangle BAE$ (c.g.c)

b)

Ta có:

$\widehat{DBC}=\widehat{ABC}+\widehat{DBA}$

$\widehat{ABF}=\widehat{ABC}+\widehat{CBF}$

mả $\widehat{DBA}=\widehat{CBF}=60^0$

$\Rightarrow \widehat{DBC}=\widehat{ABF}$

Xét $\triangle DBC$ và $\triangle ABF$ có:

$BD=AB$ (do $\triangle ABD$ đều-gt)

$BC=BF$ (do $\triangle BCF$ đều -gt)

$\widehat{DBC}=\widehat{ABF}$ (cmt)

$\Rightarrow \triangle DBC=\triangle ABF$ (c.g.c)

$\Rightarrow DC=AF$

mà $DC=BE$ (do $\triangle DAC=\triangle BAE$ -cmt)

$\Rightarrow DC=BE=AF$

c)

Do $\triangle DAC=\triangle BAE$ (cmt)

$\Rightarrow \widehat{ACD}=\widehat{AEB}$ (hai góc tương ứng)

Ta có $\widehat{AEC}+\widehat{ACE}=60^0+60^0=120^0$

$\Rightarrow \widehat{AEB}+\widehat{OEC}+\widehat{ACE}=120^0$

mà $\widehat{ACD}=\widehat{AEB}$ (cmt)

$\Rightarrow \widehat{ACD} +\widehat{OEC}+\widehat{ACE}=120^0$

$\Rightarrow \widehat{OEC}+\widehat{OCE}=120^0$

$\Rightarrow \widehat{COE}=60^0$

d)

Do $\triangle DAC=\triangle BAE$ (cmt) 

$\Rightarrow \widehat{ADC}=\widehat{ABE}$ (hai góc tương ứng)

Do $\triangle DBC=\triangle ABF$  (cmt)

$\Rightarrow \widehat{BDC}=\widehat{BAF}$ (hai góc tương ứng)

Ta có $\widehat{AOB}=180^0-\widehat{BAF}-\widehat{ABE}$

mà $\widehat{ADC}=\widehat{ABE}$ (cmt) và $ \widehat{BDC}=\widehat{BAF}$ (cmt)

$\Rightarrow \widehat{AOB}=180^0-\widehat{BDC}-\widehat{ADC}$

$\Rightarrow \widehat{AOB}=180^0-\widehat{ADB}$

$\Rightarrow \widehat{AOB}=180^0-60^0=120^0$