Question

Answer 1

\(a,AC//Oy\Rightarrow AC\perp Ox;BC//Ox\Rightarrow BC\perp Oy\\ \widehat{AOB}+\widehat{A}+\widehat{B}+\widehat{ACB}=360^0\\ \Rightarrow\widehat{ACB}=360^0-90^0-90^0-90^0=90^0\\ b,AC//BO\Rightarrow\widehat{BOD}=\widehat{ADO}\left(so.le.trong\right)\)

Mà \(\widehat{BOD}=\widehat{AOD}\left(OD.là.p/g.\widehat{xOy}\right)\)

Do đó \(\widehat{ADO}=\widehat{AOD}\)

\(c,\widehat{ACB}=\widehat{AOB}=90^0\Rightarrow\dfrac{1}{2}\widehat{ACB}=\dfrac{1}{2}\widehat{AOB}\\ \Rightarrow\widehat{ECD}=\widehat{DOA}\left(OE.là.p/g.\widehat{ACB}\right)\)

Mà \(\widehat{AOD}=\widehat{ADO}\Rightarrow\widehat{ECD}=\widehat{ADO}\)

Mà 2 góc này ở vị trí đồng vị nên CE//OD

\(d,\) Vì OD là p/g góc xOy nên \(\widehat{AOD}=\dfrac{1}{2}\widehat{AOB}=\dfrac{1}{2}\cdot90^0=45^0\)

Do đó \(\widehat{ADO}=\widehat{AOD}=45^0\) (cm ở câu a)