Đáp án:

1)

$\begin{array}{l}
a)\widehat A + \widehat B + \widehat C + \widehat D = {360^0}\\
Do:\dfrac{{\widehat A}}{4} = \dfrac{{\widehat B}}{3} = \dfrac{{\widehat C}}{2} = \dfrac{{\widehat D}}{1} = \dfrac{{\widehat A + \widehat B + \widehat C + \widehat D}}{{4 + 3 + 2 + 1}}\\
 = \dfrac{{{{360}^0}}}{{10}} = {36^0}\\
 \Rightarrow \left\{ \begin{array}{l}
\widehat A = {144^0}\\
\widehat B = {108^0}\\
\widehat C = {72^0}\\
\widehat D = {36^0}
\end{array} \right.\\
b) + )\widehat {ECD} + \widehat {EDC} + \widehat {DEC} = {180^0}\\
 \Rightarrow \dfrac{1}{2}.\widehat C + \dfrac{1}{2}.\widehat D + \widehat {DEC} = {180^0}\\
 \Rightarrow \widehat {DEC} = {180^0} - {36^0} - {18^0} = {126^0}\\
 + )\widehat {FCD} = \dfrac{1}{2}.\left( {{{180}^0} - \widehat C} \right) = {54^0}\\
\widehat {FDC} = \dfrac{1}{2}.\left( {{{180}^0} - \widehat D} \right) = {72^0}\\
 \Rightarrow \widehat {DFC} = {180^0} - {54^0} - {72^0} = {54^0}\\
2)\\
a)AB//CD\\
 \Rightarrow \widehat A + \widehat D = {180^0}\\
 \Rightarrow \widehat A = {120^0}\\
b)\dfrac{{\widehat B}}{{\widehat C}} = \dfrac{4}{5}\\
 \Rightarrow \dfrac{{\widehat B}}{4} = \dfrac{{\widehat C}}{5} = \dfrac{{\widehat B + \widehat C}}{{4 + 5}} = \dfrac{{{{180}^0}}}{9} = {20^0}\\
 \Rightarrow \left\{ \begin{array}{l}
\widehat B = {80^0}\\
\widehat C = {100^0}
\end{array} \right.\\
3)\\
\widehat {ABC} = \widehat {ACB} = {45^0}\\
\widehat {ACD} = \widehat {ADC} = {45^0}\\
 \Rightarrow \left\{ \begin{array}{l}
\widehat {ACB} + \widehat {ACD} = \widehat {BCD} = {90^0}\\
\widehat {CBA} = \widehat {CDA} = {45^0}
\end{array} \right.
\end{array}$

=> 4 điểm A,B,C,D tạo thành 1 tam giác vuông cân tại C.