Vì AD//BC nên \(\left\{{}\begin{matrix}\widehat{A}+\widehat{B}=180^0\\\widehat{C}+\widehat{D}=180^0\end{matrix}\right.\left(trong.cùng.phía\right)\Rightarrow\left\{{}\begin{matrix}\widehat{B}=90^0\\\widehat{C}=120^0\end{matrix}\right.\left(D\right)\)
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