a) Ta có: OK là tia phân giác của \(\widehat{O}\)
\(\Rightarrow\widehat{O_1}=\widehat{O_2}\)
Xét \(\Delta OKB\) và \(\Delta OK\text{A}\) có:
\(\left\{{}\begin{matrix}\widehat{OKB}=\widehat{OK\text{A}}\left(gt\right)\\OK\\\widehat{O_1}=\widehat{O_2}\left(gt\right)\end{matrix}\right.\)là cạnh chung (gt)
\(\Rightarrow\Delta OKB=\Delta OK\text{A}\left(g-c-g\right)\)
Vì \(\Delta OKB=\Delta OK\text{A}\)
\(\Rightarrow KB=K\text{A}\) (hai góc tương ứng)
b) Ta có: \(\widehat{OKB}+\widehat{OK\text{A}}=180^o\)
\(\Rightarrow\widehat{OKB}=\widehat{OK\text{A}}=180^o:2=90^o\)
Vì \(\widehat{OKB}=\widehat{OK\text{A}}\)
\(\Rightarrow OK\perp AB\)
