a) *Xét \(\Delta OAD\) và \(\Delta OCB\) có:
\(\left\{{}\begin{matrix}\widehat{O}.l\text{à}.g\text{óc}.chung\\OC=OA\left(gt\right)\\OD=OB\left(gt\right)\end{matrix}\right.\)
\(\Rightarrow\Delta OAD=\Delta OCB\left(c-c-c\right)\)
\(\Rightarrow AD=CB\) (hai cạnh tương ứng)
b) Vì \(\Delta OAD=\Delta OCB\left(cmt\right)\)
\(\Rightarrow\widehat{D}=\widehat{B}\) (hai góc tương ứng)
*Ta có: \(\left\{{}\begin{matrix}OD=OB\\OC=OA\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}CD=OD-OC\\AB=OB-OA\end{matrix}\right.\)
\(\Rightarrow CD=AB\)
*Xét \(\Delta EAB\) và \(\Delta ECD\) có:
\(\left\{{}\begin{matrix}\widehat{AEB}=\widehat{CED}\left(hai.g\text{óc}.\text{đ}\text{ối}.\text{đ}\text{ỉnh}\right)\\AB=CD\left(cmt\right)\\\widehat{B}=\widehat{D}\left(cmt\right)\end{matrix}\right.\)
\(\Rightarrow\Delta EAD=\Delta ECD\left(g-c-g\right)\)
c) *Vì \(\Delta EAB=\Delta ECD\left(cmt\right)\)
\(\Rightarrow EA=EC\) (hai cạnh tương ứng)
*Xét \(\Delta OEC\) và \(\Delta OEA\) có:
\(\left\{{}\begin{matrix}OC=OA\left(gt\right)\\EC=EA\left(cmt\right)\\OE.l\text{à}.c\text{ạnh}.chung\end{matrix}\right.\)
\(\Rightarrow\Delta OEC=\Delta OEA\left(c-c-c\right)\)
\(\Rightarrow\widehat{COE}=\widehat{AO\text{E}}\) (hai góc tương ứng)
*Ta có: \(\left\{{}\begin{matrix}\widehat{COE}=\widehat{AO\text{E}}\left(cmt\right)\\OE.n\text{ằm}.gi\text{ữa}.OA.v\text{à}.OC\end{matrix}\right.\)
\(\Rightarrow OE\) là tia phân giác của góc xOy
