Gọi \(J=CE\cap AB\), \(F=BD\cap AC\) , \(H=CE\cap BD\)
Có \(\widehat{EAB}=\widehat{ECB}=\dfrac{1}{2}sđ\stackrel\frown{EB}\)
\(\widehat{CAD}=\widehat{DBC}=\dfrac{1}{2}sđ\stackrel\frown{DC}\)
\(\Rightarrow\widehat{EAB}+\widehat{CAD}=\widehat{ECB}+\widehat{DBC}=180^0-\widehat{BHC}\) (*)
Lại có \(\widehat{AJC}+\widehat{AFB}=180^0\) => Tứ giác AJHF nội tiếp đường tròn
\(\Rightarrow180^0=\widehat{BAC}+\widehat{JHF}=\widehat{BAC}+\widehat{BHC}\)
\(\Rightarrow180^0-\widehat{BHC}=\widehat{BAC}\) (2*)
Từ (*); (2*) => \(\widehat{EAB}+\widehat{CAD}=\widehat{BAC}\)
\(\Leftrightarrow\widehat{EAB}+\widehat{BAC}+\widehat{CAD}=2\widehat{BAC}\)
\(\Leftrightarrow\widehat{EAD}=2\alpha\)
Ý C
