Đáp án:
$\widehat{A}=45^\circ,\widehat{B}=115^\circ,\widehat{C}=65^\circ,\widehat{D}=135^\circ$
Giải thích các bước giải:
$AB//CD\\ \Rightarrow \widehat{A}+\widehat{D}=180^\circ ,\widehat{B}+\widehat{C}=180^\circ\\ \widehat{A}+\widehat{D}=180^\circ\\ \Leftrightarrow \dfrac{1}{3}\widehat{D}+\widehat{D}=180^\circ\\ \Leftrightarrow \dfrac{4}{3}\widehat{D}=180^\circ\\ \Leftrightarrow \widehat{D}=135^\circ\\ \Rightarrow \widehat{A}=180^\circ-135^\circ=45^\circ\\ \widehat{B}+\widehat{C}=180^\circ(1)\\ \widehat{B}-\widehat{C}=50^\circ(2)$
Cộng 2 vế $(1)(2)$ ta có $2\widehat{B}=230^\circ \Leftrightarrow \widehat{B}=115^\circ$
$\Rightarrow \widehat{C}=180^\circ-\widehat{B}=65^\circ$
