Đáp án:
$\begin{array}{l}
Do:Ax//By\\
\Rightarrow \widehat {CAx} = \widehat {CDB} = {50^0}\left( {so\,le\,trong} \right)\\
Trong:\Delta BCD:\\
\widehat {CBy} + \widehat {CDB} + \widehat {BCD} = {180^0}\\
\Rightarrow \widehat {BCD} = {180^0} - {50^0} - {40^0} = {90^0}\\
Do:\widehat {ACB} + \widehat {BCD} = \widehat {ACD} = {180^0}\\
\Rightarrow \widehat {ACB} = {180^0} - {90^0} = {90^0}\\
\text{Vậy}\,\widehat {CDB} = {50^0};\widehat {ACB} = {90^0};\widehat {BCD} = {90^0}
\end{array}$