Theo tc 2 tt cắt nhau: \(MC=AC;MD=BD\)
\(\left\{{}\begin{matrix}\widehat{CAO}=\widehat{CMO}=90^0\\AC=CM\\CO.chung\end{matrix}\right.\Rightarrow\Delta ACO=\Delta MCO\left(ch-cgv\right)\\ \Rightarrow\widehat{AOC}=\widehat{MOC}=\dfrac{1}{2}\widehat{AOM}\\ \left\{{}\begin{matrix}\widehat{OMD}=\widehat{OBD}=90^0\\MD=BD\\OD.chung\end{matrix}\right.\Rightarrow\Delta BDO=\Delta MDO\left(ch-cgv\right)\\ \Rightarrow\widehat{BOD}=\widehat{MOD}=\dfrac{1}{2}\widehat{BOM}\)
Ta có \(\widehat{COD}=\widehat{COM}+\widehat{DOM}=\dfrac{1}{2}\left(\widehat{BOM}+\widehat{AOM}\right)=\dfrac{1}{2}\widehat{AOB}=\dfrac{1}{2}\cdot180^0=90^0\)
Vậy DOC vuông tại O
