anh làm tắt nhekVẽ $EH$ vuông $AB(H\in AB)$
$\triangle HAE\sim \triangle DAB(g-g)\Rightarrow \frac{AE}{AB}=\frac{AH}{AD}\Rightarrow AE.AD=AB.AH$
$\triangle HEB\sim \triangle CAB(g-g)\Rightarrow \frac{HB}{BC}=\frac{BE}{AB}\Rightarrow BE.BC=AB.HB$
$\Rightarrow AE.AD+BE.BC=AB.AH+AB.HB=AB(AH+HB)=AB^2=(2R)^2=4R^2$
Xong rùi