Cho tam giác ABC nội tiếp trong đường tròn tâm O

Xét đường tròn (O) có:

$\widehat{A}=\frac{1}{2}$sđ$\stackrel{⏜}{BC}$(tính chất góc nội tiếp)

$\Rightarrow$sđ$\stackrel{⏜}{BC}=2\widehat{A}={2.32}^o={64}^o$

Ta có: BC = BE (gt)

$\Rightarrow$sđ$\stackrel{⏜}{BC}$= sđ$\stackrel{⏜}{BE}={64}^o$

Mà $\widehat{B}=\frac{1}{2}$sđ$\stackrel{⏜}{AC}$(tính chất góc nội tiếp)

$\Rightarrow$sđ$\stackrel{⏜}{AC}=2\widehat{B}={2.84}^o={168}^o$

Lại có: AC = CF (gt)

$\Rightarrow$sđ$\stackrel{⏜}{CF}$= sđ$\stackrel{⏜}{AC}={168}^o$

sđ$\stackrel{⏜}{AC}$+ sđ$\stackrel{⏜}{AF}$+ sđ$\stackrel{⏜}{CF}$=${360}^o$

$\Rightarrow$sđ$\stackrel{⏜}{AF}$=${360}^o$ - sđ$\stackrel{⏜}{AC}$- sđ$\stackrel{⏜}{CF}$=${360}^o-{168}^o.2={24}^o$

Xét tam giác ABC có:

$\widehat{A}+\widehat{B}+\widehat{C}={180}^o$

$\Rightarrow \widehat{ACB}={180}^o-\left(\widehat{A}+\widehat{B}\right)={180}^o-\left({32}^o+{84}^o\right)={64}^o$

Mà $\widehat{ACB}=\frac{1}{2}$sđ$\stackrel{⏜}{AB}$(góc nội tiếp chắn cung)

$\Rightarrow$sđ$\stackrel{⏜}{AB}=2\widehat{ACB}={2.64}^o={128}^o$

Lại có: AD = AB (gt)

$\Rightarrow$sđ$\stackrel{⏜}{AD}$= sđ$\stackrel{⏜}{AB}={128}^o$

Ta có: $\widehat{FED}=\frac{1}{2}$sđ$\stackrel{⏜}{DF}=\frac{1}{2}$(sđ$\stackrel{⏜}{AD}$+ sđ$\stackrel{⏜}{AF}$) =$\frac{1}{2}\left({128}^o+{24}^o\right)={76}^o$

$\widehat{EDF}=\frac{1}{2}sd\stackrel{⏜}{EF}=\frac{1}{2}\left(sd\stackrel{⏜}{AB}-sd\stackrel{⏜}{AF}-sd\stackrel{⏜}{BE}\right)=\frac{1}{2}\left({128}^o-{24}^o-{64}^o\right)={20}^o$

$\widehat{DFE}={180}^o-\left(\widehat{FED}+\widehat{EDF}\right)={180}^o-\left({76}^o+{20}^o\right)={84}^o$