Theo đề ra, ta có: \(\widehat{x'Oy'}\)và \(\widehat{xOy}\)đối nhau
\(\rightarrow\widehat{xOy}=\widehat{x'Oy'}\)\(\Rightarrow\frac{1}{2}\)\(\widehat{xOy}=\frac{1}{2}\widehat{x'Oy'}\)
\(\rightarrow\widehat{xOa}=\widehat{yOb}\)
Ta có: \(\widehat{xOy}\)thẳng hàng
\(\Rightarrow Oa\)và \(Ob\)đối đỉnh