Cho hình vẽ : 
ADBxC

Biết $\widehat{A}=120^0$; $\widehat{C}=30^0$; $\widehat{D}=60^0$
1/ Chứng minh rằng : $AB // DC$.
2/ Tính $\widehat{ABC}$ và $\widehat{xBC}$ ?
2/
$AB//CD\Rightarrow \widehat{ABC}+\widehat{ACB}=180^0$
$\Rightarrow\widehat{ABC}=180^0-\widehat{ACB}=180^0-30^0=150^0$
$\widehat{xBC}=\widehat{BCD}=30^0$
sửa từng chỗ 1 –  ❄⊰๖ۣۜNgốc๖ۣۜ ⊱ ❄ 19-08-16 10:17 AM
hiện nó đang loạn hết cả lên, chưa sửa đc –  Effort 19-08-16 10:07 AM
sửa lại đề đi Quân –  ❄⊰๖ۣۜNgốc๖ۣۜ ⊱ ❄ 19-08-16 10:06 AM
a) Có $\widehat{A}+\widehat{D}=120^{o}+60^o=180^o$
Mà $\widehat{A};\widehat{D}$ là 2 góc trong cùng phía
Nên $AB//CD$
Ta có: $\widehat{A}+\widehat{D}=180^0$. Mà 2 góc nằm ở vị trí trong cùng phía
$\Rightarrow AB//CD$
a) Có $\widehat{A}+\widehat{D}=120^{o}+60^o=180^o$
Mà $\widehat{A};\widehat{D}$ là 2 góc trong cùng phía
Nên $AB//CD$
b) có \widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^{0}
thay số 120^{0}+\widehat{B}+30^{0}+60^{0}=360^{0}

\widehat{B}=150^{0}
ta có \widehat{ABC}+\widehat{zBC}=180^{0}(2 góc kề bù)
thay số 150^{0}+\widehat{zBC}=180^{0}
\widehat{zBC}=30^{0}

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