Bài 9:
Xét ΔBAC có BA=BC
nên ΔBAC cân tại B
Suy ra: \(\widehat{BAC}=\widehat{BCA}\)
mà \(\widehat{BAC}=\widehat{DAC}\)
nên \(\widehat{DAC}=\widehat{BCA}\)
mà hai góc này là hai góc ở vị trí so le trong
nên AD//BC
hay ADCB là hình thang
\(7,\)
\(AB//CD\Rightarrow\left\{{}\begin{matrix}\widehat{A}+\widehat{D}=180^0\\\widehat{B}+\widehat{C}=180^0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2\widehat{D}+40^0=180^0\\3\widehat{C}=180^0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}\widehat{D}=70^0\\\widehat{C}=60^0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\widehat{A}=110^0\\\widehat{B}=120^0\end{matrix}\right.\)
\(10,\)
\(a,\widehat{D}-\widehat{C}=20^0\Rightarrow\widehat{D}=20^0+\widehat{C}=20^0+3\widehat{A}\)
Vì ABCD là tứ giác nên \(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^0\)
\(\Rightarrow\widehat{A}+\widehat{A}+20^0+3\widehat{A}+20^0+3\widehat{A}=360^0\\ \Rightarrow8\widehat{A}=320^0\Rightarrow\widehat{A}=40^0\\ \Rightarrow\left\{{}\begin{matrix}\widehat{B}=40^0+20^0=60^0\\\widehat{C}=3\cdot40^0=120^0\\\widehat{D}=20^0+3\cdot40^0=140^0\end{matrix}\right.\)
