Chứng minh OE=OK=OD biết tam giác ABC có góc B=60 độ, tia phân giác AD

a) \(\Delta ABC\) có \(\widehat{BAC}+\widehat{B}+\widehat{ACB}=180^o\)

\(\Rightarrow\widehat{BAC}+\widehat{ACB}=120^o\)

\(\Rightarrow\widehat{\dfrac{BAC}{2}}+\dfrac{\widehat{ACB}}{2}=\dfrac{120^o}{2}=60^o\)

\(\Rightarrow\widehat{OAC}+\widehat{OCA}=60^o\) (vì AD là tia phân giác của \(\widehat{BAC}\), CE là tia phân giác của \(\widehat{ACB}\))

\(\Delta AOC\) có: \(\widehat{AOC}+\widehat{OAC}+\widehat{OCA}=180^o\)

\(\Rightarrow\widehat{AOC}=120^o\)

b) Vì \(\widehat{AOC}+\widehat{AOE}=180^o\)(kề bù) nên \(\widehat{AOE}=60^o\)

\(\Rightarrow\widehat{AOE}=\widehat{COD}=60^o\) (đối đỉnh)

\(\Delta EOA\) và \(\Delta KOA\) có: \(\left\{{}\begin{matrix}AE=AK\\\widehat{EAO}=\widehat{KAO}\\AOchung\end{matrix}\right.\)

\(\Rightarrow\)\(\Delta EOA\)= \(\Delta KOA\)(c.g.c)

\(\Rightarrow\left\{{}\begin{matrix}OE=OK\\\widehat{AOE}=\widehat{AOK}=60^o\end{matrix}\right.\)

Ta có: \(\widehat{AOE}+\widehat{AOK}+\widehat{COK}=180^o\)

\(\Rightarrow\widehat{COK}=60^o\)

\(\Delta KOC\) và \(\Delta DOC\) có: \(\left\{{}\begin{matrix}\widehat{COK}=\widehat{COD}\left(=60^o\right)\\OCchung\\\widehat{KOC}=\widehat{DOC}\end{matrix}\right.\)

\(\Rightarrow\) \(\Delta KOC\)= \(\Delta DOC\) (g.c.g)

\(\Rightarrow OK=OD\)

Vì OK = OE, OK= OD nên OK=OD=OE