a) Xét Tam giác AOB có:
\(\widehat{A}+\widehat{B}+\widehat{C}=180^o\Rightarrow\widehat{B}+\widehat{C}=180^o-\widehat{A}\)
Xét tam giác BOC có:\(\widehat{B_1}+\widehat{C_1}+\widehat{BOC}=180^o\Rightarrow\widehat{B_1}+\widehat{C_1}=180^o-\widehat{BOC}\)
Mà \(\widehat{B_1}=\widehat{B_2}=\frac{1}{2}\widehat{B}\)(BD là phân giác )
\(\widehat{C_1}=\widehat{C_2}=\frac{1}{2}\widehat{C}\)
\(\Rightarrow2\widehat{B_1}+2\widehat{C_1}=180^o-\widehat{A}\Rightarrow2\left(\widehat{B_1}+\widehat{C_1}\right)=180^o-\widehat{A}\)\(\Rightarrow2\left(180^o-\widehat{BOC}\right)+\widehat{A}=180^o\Rightarrow\widehat{BOC}=90^o+\frac{\widehat{A}}{2}=90^o+120^o:2=150^o\)
\(\Rightarrow\widehat{IOK}=\widehat{BOC}-\widehat{BOI}-\widehat{KOC}=150^o-30^o-30^o=90^o\)
=> OI vông OK
b)Ta có:
\(\widehat{EOB}=\widehat{DOC}=180^o-\widehat{BOC}=30^o\)
Xét tam giác EBO và IBO có:
BO chung
\(\widehat{B_1}=\widehat{B_2}\)( phân giác )
\(\widehat{BOE}=\widehat{BOI}=30^o\)
=> \(\Delta BEO=\Delta BIO\)(g.c.g)
=> BE=BI
Tương tự ta chứng minh đc: \(\Delta CDO=\Delta CKO\)(g.c.g)=> CD=CK
Mà BI+IK+KC=BC=> BE+IK+CD=BC
=> BE+CD< BC