Ta có hình vẽ :
Ta có : \(\left\{{}\begin{matrix}\widehat{B}-\widehat{D}=10^o\\\widehat{A}-\widehat{B}=21^o\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\widehat{B}=\widehat{D}+10^o\\\widehat{A}=21^o+\widehat{B}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\widehat{B}=\widehat{D}+10^o\\\widehat{A}=21^o+\widehat{D}+10^o=31^o+\widehat{D}\end{matrix}\right.\)
\(\Delta ADC\) cân tại D ( AD = DC ) \(\Rightarrow\widehat{DAC}=\widehat{DCA}\left(1\right)\)
\(\Delta ABC\) cân tại B ( AB = BC ) \(\Rightarrow\widehat{BAC}=\widehat{BCA}\left(2\right)\)
Từ ( 1 ) ; ( 2 ) \(\Rightarrow\widehat{DAC}+\widehat{BAC}=\widehat{DCA}+\widehat{BCA}\)
\(\Rightarrow\widehat{A}=\widehat{C}\)
Mà \(\widehat{A}=31^o+\widehat{D}\Rightarrow\widehat{C}=31^o+\widehat{D}\)
Xét tứ giác ABCD có :
\(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^o\)
\(\Rightarrow31^o+\widehat{D}+10^o+\widehat{D}+31^o+\widehat{D}+\widehat{D}=360^o\)
\(\Rightarrow4\widehat{D}+72^o=360^o\)
\(\Rightarrow4\widehat{D}=288^o\)
\(\Rightarrow\widehat{D}=72^o\)
\(\Rightarrow\left\{{}\begin{matrix}\widehat{B}=72^o+10^o=82^o\\\widehat{A}=\widehat{C}=72^o+31^o=103^o\end{matrix}\right.\)
Vậy \(\widehat{A}=\widehat{C}=103^o;\widehat{B}=82^o;\widehat{D}=72^o\)