Question

Lm gấp giúp em ạ, 14h50 e nộp

Answer 1

\(\left\{{}\begin{matrix}\widehat{AOD}+\widehat{BOC}=180^0\\\widehat{AOD}=\widehat{BOC}\left(đối.đỉnh\right)\end{matrix}\right.\Rightarrow\widehat{AOD}=\widehat{BOC}=90^0\\ \Rightarrow AB\perp CD\\ \Rightarrow\widehat{AOD}=\widehat{BOD}=\widehat{BOC}=\widehat{AOC}=90^0\)

Answer 2

Ta có: \(\widehat{AOD}=\widehat{BOC}\)(đối đỉnh)

Mà \(\widehat{AOD}+\widehat{BOC}=100^0\)

\(\Rightarrow\widehat{AOD}=\widehat{BOC}=50^0\)

Ta có: \(\widehat{AOD}+\widehat{AOC}=180^0\)(kề bù) 

\(\Rightarrow\widehat{AOC}=180^0-\widehat{AOD}=130^0\)

\(\Rightarrow\widehat{DOB}=\widehat{AOC}=130^0\)(đối đỉnh)