\(a,\widehat{OBA}+\widehat{OCA}=90^0+90^0=180^0\\ \Rightarrow ABOC\text{ nội tiếp}\\ \RightarrowĐpcm\\ b,\left\{{}\begin{matrix}AB=AC\Rightarrow A\in\text{trung trực }BC\\OB=OC=R\Rightarrow O\in\text{trung trực }BC\end{matrix}\right.\\ \Rightarrow OA\text{ là trung trực }BC\\ c,D\text{ đx }B\text{ qua }O\Rightarrow OD=OB=R\\ \Rightarrow D\in\left(O;R\right)\\ \Rightarrow\widehat{BED}=90^0\\ \left\{{}\begin{matrix}\widehat{BED}=\widehat{ABD}=90^0\\\widehat{ABD}\text{ chung}\end{matrix}\right.\Rightarrow\Delta BED\sim\Delta ABD\left(g.g\right)\\ \Rightarrow\dfrac{DE}{BE}=\dfrac{BD}{BA}\\ d,\left\{{}\begin{matrix}\widehat{BCD}=\widehat{AHB}=90^0\\\widehat{ABH}=\widehat{BDC}\left(\text{cùng phụ }\widehat{BCD}\right)\end{matrix}\right.\Rightarrow\Delta BCD\sim\Delta AHB\left(g.g\right)\)
\(\Rightarrow\dfrac{BD}{AB}=\dfrac{CD}{HB}\Rightarrow\dfrac{DE}{BE}=\dfrac{CD}{HB}\left(=\dfrac{BD}{BA}\right)\)
Lại có \(BECD\) nội tiếp nên \(\widehat{EBC}=\widehat{EDC}\)
Do đó \(\Delta BHE\sim\Delta DCE\left(c.g.c\right)\)
\(\Rightarrow\widehat{BEH}=\widehat{CED}\\ \Rightarrow\widehat{BEH}+\widehat{HED}=\widehat{CED}+\widehat{HED}\\ \Rightarrow\widehat{HEC}=\widehat{BED}=90^0\)
