Xét đường tròn \((O)\) có:
\(\widehat A =\displaystyle {1 \over 2} sđ \overparen{BC}\) (tính chất góc nội tiếp)
\( \Rightarrow sđ \overparen{BC}\) \( = 2\widehat A = {2.32^o} = {64^o}\)
Ta có: \(BC = BE \;\;(gt)\)
\( \Rightarrow sđ \overparen{BC}\)\( = sđ \overparen{BE}= 64^o\)
Mà \(\widehat B = \displaystyle {1 \over 2} sđ \overparen{AC}\) (tính chất góc nội tiếp)
\( \Rightarrow \) sđ \(\overparen{AC}\) \( = 2\widehat B = {2.84^o} = {168^o}\)
Lại có: \(AC = CF \;\;(gt)\)
\( \Rightarrow sđ \overparen{CF}\) \(= sđ \overparen{AC}= 168^o\)
\( sđ \overparen{AC} + sđ \overparen{AF} + sđ \overparen{CF}\)\( = 360^o\)
\( \Rightarrow sđ \overparen{AF}\) \( = {360^o} - sđ \overparen{AC} - sđ \overparen{CF}\)\( = 360^o – 168^o. 2 = 24^o\)
Trong \(∆ABC\) ta có: \(\widehat A + \widehat B + \widehat C = {180^o}\)
\( \Rightarrow \widehat {ACB} = {180^0} - \left( {\widehat A + \widehat B} \right)\)
\( = {180^0} - \left( {{{32}^o} + {{84}^o}} \right) = {64^o}\)
Mà \( \widehat {ACB} = \displaystyle {1 \over 2} sđ \overparen{AB}\)
\( \Rightarrow sđ \overparen{AB} = 2\widehat {ACB} = {2.64^o} = {128^o}\)
Lại có \(AD = AB\;\; (gt)\)
\( \Rightarrow sđ \overparen{AD} = sđ \overparen{AB} = 128^o\)
Ta có: \(\widehat {FED} = \displaystyle {1 \over 2} sđ \overparen{DF}\) \( =\displaystyle {1 \over 2} ( sđ \overparen{AD} + sđ \overparen{AF}\))
\(= \displaystyle{1 \over 2}.\left( {{{128}^o} + {{24}^o}} \right) = {76^o}\)
\(\widehat {EDF} = \displaystyle{1 \over 2} sđ \overparen{EF}\) \(=\displaystyle {1 \over 2} ( sđ \overparen{AB} - sđ \overparen{AF} - sđ \overparen{BE})\)
\(= \displaystyle{1 \over 2}.\left( {{{128}^o} - {{24}^o} - {{64}^o}} \right) = {20^o}\)
\(\widehat {DFE} = {180^o} - \left( {\widehat {FED} + \widehat {EDF}} \right)\)
\(= {180^0} - \left( {{{76}^o} + {{20}^o}} \right) = {84^o}\).
