Ta có: AD//BC(ABCD là hthang cân)
\(\Rightarrow\left\{{}\begin{matrix}\widehat{A}+\widehat{B}=180^0\left(trong.cùng.phía\right)\\\widehat{A}-\widehat{B}=40^0\left(gt\right)\end{matrix}\right.\)
\(\Rightarrow\widehat{A}=\left(180^0+40^0\right):2=110^0\)
\(\Rightarrow\widehat{D}=\widehat{A}=110^0\)(ABCD là hthang cân)