Xé đường tròn \((O),\) ta có: \(\widehat{MFE}= \dfrac{1}{2}sđ\overparen{ME}(nhỏ)\)
\(=\dfrac{1}{2}(sđ \overparen{PM} +sđ \overparen{PE})\)
\(\widehat{N}=\dfrac{1}{2}(sđ \overparen{EF} -sđ \overparen{PM})\)
\(\Rightarrow sđ \overparen{EF}=2\widehat{N}+ sđ \overparen{PM}\)
\(\widehat{Q}=\dfrac{1}{2}(sđ \overparen{MF} -sđ \overparen{PE})\)
\(\Rightarrow sđ \overparen{MF}=2 \widehat{Q}+sđ \overparen{PE}\)
\(\Rightarrow sđ \overparen{MF}+sđ\overparen{EF}\)\(=2 (\widehat{Q}+\widehat{N})+(sđ \overparen{PE}+sđ \overparen{PM})\)
\(\Rightarrow sđ \overparen{ME}(lớn)\)\(= 2 (\widehat{Q}+\widehat{N})+sđ \overparen{ME}(nhỏ)\)
Mà \(sđ \overparen{ME}(lớn)=360^\circ-sđ \overparen{ME}(nhỏ)\)
\(\Rightarrow 360^\circ-sđ \overparen{ME}(nhỏ)\)\(= 2 (\widehat{Q}+\widehat{N})+sđ \overparen{ME}(nhỏ)\)
\(\Rightarrow sđ \overparen{ME}(nhỏ) =180^\circ-(\widehat{Q}+\widehat{N})\)\(=180^\circ-(35^\circ+45^\circ)=100^\circ\)
Do đó: \(\widehat{MFE}= \dfrac{1}{2}sđ\overparen{ME}(nhỏ)\)\(= \dfrac{1}{2}.100^\circ=50^\circ\)
Vậy chọn \((A)\) \(50^\circ\)
