a) Ta có: \(\widehat{IOK}=\widehat{BOC}-\widehat{BOI}-\widehat{KOC}=\widehat{BOC}-60^o\)
Mà \(\widehat{BOC}=180^o-\widehat{B_1}-\widehat{C_1}=180^o-\left(\frac{\widehat{B}}{2}+\frac{\widehat{C}}{2}\right)=180^o-\frac{180^o-\widehat{A}}{2}=180^o-30^o=150^o\)
\(\Rightarrow\widehat{IOK}=150^o-60^o=90^o\Rightarrow OI\perp OK\)
b) Ta có: \(\widehat{BOE}=\widehat{COD}=180^o-30^o-90^o-30^o=30^o\)
Xét \(\Delta BEO;\Delta BIO\); có:
\(\widehat{B_1}=\widehat{B_2}\left(gt\right);\) Chung BO \(;\widehat{IOB}=\widehat{EOB}=30^o\)
=> \(\Rightarrow\Delta BEO=\Delta BIO\left(g.c.g\right)\Rightarrow BE=BI.\)
Tương tự thì KC=DC
Mà BC>BI+KC => BE > BE+DC
