Ta có : \(\widehat{A}-\widehat{D}=20^o\Rightarrow\widehat{A}=20^o+\widehat{D}\)
\(AB//CD\Rightarrow\widehat{A}+\widehat{D}=180^o\)
( Hai góc trong cùng phía bù nhau )
\(\Rightarrow20^o+\widehat{D}+\widehat{D}=180^o\)
\(\Leftrightarrow2.\widehat{D}=160^o\)
\(\Leftrightarrow\widehat{D}=80^o\)
\(\Rightarrow\widehat{A}=\widehat{D}+20^o=100^o\)
Từ đó , ta lại có : \(AB//CD\Rightarrow\widehat{B}+\widehat{C}=180^o\)
( Hai góc trong cùng phía bù nhau )
Mà \(\widehat{B}=2\widehat{C}\Rightarrow2\widehat{C}+\widehat{C}=180^o\)
\(\Rightarrow3\widehat{C}=180^o\)
\(\Leftrightarrow\widehat{C}=60^o\)
\(\Leftrightarrow\widehat{B}=2.\widehat{C}=120^o\)
Vậy : \(\widehat{A}=100^o\)
\(\widehat{B}=120^o\)
\(\widehat{C}=60^o\)
\(\widehat{D}=80^o\)
