Giải thích các bước giải:

Vì $\widehat{BOC}=120^o$

$\rightarrow\widehat{BAC}=60^o$

$\rightarrow\widehat{BAE}=\widehat{BAC}=\widehat{CAD}=60^o$

$\rightarrow A,D,E $ thẳng hàng

Lại có:

$\widehat{ABK}+\widehat{ADC}=180^o$

$\widehat{ADC}=\widehat{ABC}=\widehat{ABE}$

$\rightarrow\widehat{ABE}+\widehat{ABK}=180^o\rightarrow E,B,K$ thẳng hàng

Chứng minh tương tự ta có D,C,K thẳng hàng

Ta có:

$\widehat{BAK}=\widehat{BDK}=\widehat{DBC}=90^o-\widehat{ACB}=\widehat{BAO}$

$\rightarrow A,O,K$ thẳng hàng

Lại có:

$\widehat{EAB}=\widehat{EKD}\rightarrow\widehat{EKD}=60^o$

$\rightarrow cos\widehat{BKC}=\dfrac{1}{2}$

$\rightarrow cos\widehat{BKC}=\dfrac{KB^2+KC^2-BC^2}{2KB.KC}=\dfrac{1}{2}$

$\rightarrow KB^2+KC^2-BC^2=KB.KC$ 

$\rightarrow KB^2+KC^2=KB.KC+BC^2\ge 2KB.KC\rightarrow KB.KC\le BC^2$ 

$\rightarrow \dfrac{1}{KB^2}+\dfrac{1}{KC^2}\ge 2\sqrt{\dfrac{1}{KB^2}.\dfrac{1}{KC^2}}=\dfrac{2}{KB.KC}\ge \dfrac{2}{BC^2}$

Dấu = xảy ra

$\rightarrow KB=KC\rightarrow\Delta KBC$ đều

$\rightarrow KO\perp BC\rightarrow AO\perp BC$