Kẻ $OH \perp CD$
$\Delta OCD$ cân tại $O(OC=OD)$ có $OH$ vừa là đường cao vừa là đường trung tuyến, phân giác
$\Rightarrow H$ là trung điểm $CD;\widehat{HOD}=\dfrac{1}{2}\widehat{COD};\widehat{C_2}=\widehat{D_2}$
$\Rightarrow HD=\dfrac{R\sqrt{3}}{2}$
$\Delta OHD$ vuông tại $H$
$\sin\widehat{HOD}=\dfrac{HD}{OD}=\dfrac{R\sqrt{3}}{2}\\ \Rightarrow \widehat{HOD}=60^o\\ \Rightarrow \widehat{COD}=120^o;\widehat{C_2}=\widehat{D_2}=30^o(1)$
$\Delta OAC$ cân tại $O(OA=OC)$
$\Rightarrow \widehat{AOC}=180^o-2\widehat{ACO}(2)$
$\Delta OBD$ cân tại $O(OB=OD)$
$\Rightarrow \widehat{BOD}=180^o-2\widehat{BDO}(3)$
$\Delta ABO$ đều$(AO=BO=AB=R)$
$\Rightarrow \widehat{AOB}=60^o \Rightarrow \widehat{AOC}+\widehat{BOD}=\widehat{COD}-\widehat{AOB}=60^o(4)\\ (1);(2);(3);(4)\Rightarrow\widehat{AOC}+\widehat{BOD}=360^o-2\widehat{ACO}-2\widehat{BDO}\\ \Leftrightarrow \widehat{AOC}+\widehat{BOD}=360^o-2(\widehat{C_1}+\widehat{C_2}+\widehat{D_1}+\widehat{D_2})\\ \Leftrightarrow 60^o=240^o-2(\widehat{C_1}+\widehat{D_1})\\ \Leftrightarrow \widehat{C_1}+\widehat{D_1}=90^o\\ \Rightarrow \widehat{A}+\widehat{B}=360^o-\widehat{C_1}-\widehat{D_1}=270^o$
