(Hình ảnh chỉ mang tính chất minh họa)
a)* Xét \(\Delta OCB\) và \(\Delta OAD\) có:
\(\left\{{}\begin{matrix}OA=OC\left(gt\right)\\\widehat{O}.l\text{à}.g\text{óc}.chung\\OB=O\text{D}\left(gt\right)\end{matrix}\right.\)
\(\Rightarrow\Delta OCB=\Delta OAD\left(c-g-c\right)\)
\(\Rightarrow AD=BC\) (hai cạnh tương ứng)
b) *Ta có: \(\Delta OCB=\Delta OAD\left(cmt\right)\)
\(\Rightarrow\) \(\widehat{B}=\widehat{D}\) (1)
*Ta có: \(\left\{{}\begin{matrix}OA=OC\\OB=O\text{D}\end{matrix}\right.\Rightarrow AB=CD\) (2)
* Ta có: \(\Delta OCB=\Delta OAD\left(cmt\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\widehat{OAD}=\widehat{OCB}\\\widehat{OAD}+\widehat{DAB}=180^o\\\widehat{OCB}+\widehat{BCD}=180^o\end{matrix}\right.\)
\(\Rightarrow\widehat{DAB}=\widehat{BC\text{D}}\) (3)
*Từ (1), (2) và (3) \(\Rightarrow\Delta MAB=\Delta MCD\)
c) *Xét \(\Delta OAM\) và \(\Delta OCM\) có:
\(\left\{{}\begin{matrix}OA=OC\left(gt\right)\\AM=CM\\OM.l\text{à.}c\text{ạnh.}chung\end{matrix}\right.\) (Vì \(\Delta MAB=\Delta MCD\Rightarrow\) 2 góc tương ứng bằng nhau)
\(\Rightarrow\Delta OAM=\Delta OCM\left(c-c-c\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\widehat{AOM}=\widehat{COM}\left(2.g\text{óc.}t\text{ương.ứng}\right)\\OM.n\text{ằm}.gi\text{ữa}.OC.v\text{à.}O\text{A}\end{matrix}\right.\)
\(\Rightarrow OM\) là tia phân giác của \(\widehat{xOy}\)
