Ta có: a//b(gt)
\(\Rightarrow x+\widehat{BAC}=180^0\)(trong cùng phía)
\(\Rightarrow x=180^0-\stackrel\frown{BAC}=180^0-100^0=80^0\)
Ta có: a//b(gt)
\(\Rightarrow\widehat{BDC}+\widehat{ABD}=180^0\)(trong cùng phía)
\(\Rightarrow\widehat{BDC}=180^0-\widehat{ABD}=180^0-120^0=60^0\)
\(\Rightarrow y=\widehat{BDC}=60^0\)(đối đỉnh)